Answer:
atmosphere
Explanation:
.........,....
A person pushes an object of mass 5.0 kg along the floor by applying a
force. If the object experiences a friction force of 10 N and accelerates at
18 m/s^2, what is the magnitude of the force exerted by the person?
Answer:
The magnitude of the force exerted by the person is 100 N
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
Fn = ma
Where a is the acceleration of the object.
The net force is the sum of all forces exerted over a body. When an object is moved along a rough surface it experiences two horizontal forces and two vertical forces (provided there is no vertical component of the applied force).
The vertical forces are the Normal and the Weight and they are balanced, i.e.: N = W = mg.
The horizontal forces are The applied force (Fa) and the friction force (Fr). They are not balanced because the object is accelerated in that direction. The net force is:
Fn = Fa - Fr
Applying the first equation:
Fa - Fr = ma
Solving for Fa:
Fa = Fr + ma
Substituting the given values m=5 kg, Fr=10 N, [tex]a=18\ m/s^2[/tex].
Fa = 10 + 5*18 = 10 + 90 = 100
Fa = 100 N
The magnitude of the force exerted by the person is 100 N
Basically.
Given: F + m x a
Equation: 10 N + 5.0 kg x 18 m/s2
Solve: 100 N
The chain of DNA consists of
ILL MARK BRAINLIST !! get it right
What are some iron objects magnetic and some are not
Iron is magnetic, so any metal with iron in it will be attracted to a magnet. Steel contains iron, so a steel paperclip will be attracted to a magnet too. Most other metals, for example, aluminum, copper, and gold are NOT magnetic. Two metals that aren't magnetic are gold and silver.
How are wavelength and frequency related?
As wavelength increases frequency increases.
b. As wavelength decreases frequency decreases.
As wavelength increases frequency decreases.
d. None of the above
Answer:
Hey dear
Explanation:
Its option C
As Wavelength increases frequency decreases
In other case,
When Wavelength decreases frequency increases
its opposite
Tq
In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity [tex]V_{i}[/tex] = 60 mph = 26.8224 m/s
Final velocity [tex]V_{f}[/tex] = 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = [tex]\frac{1}{2}[/tex]m( [tex]V_{i}[/tex]² - [tex]V_{f}[/tex]² )
we substitute
Δk = [tex]\frac{1}{2}[/tex]×1814.37( (26.8224)² - (13.4112)² )
Δk = [tex]\frac{1}{2}[/tex] × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car? So assuming that I've sketched a graph, do I just do distance x time?
a. The truck travels twice as far as the car.
b.There is not enough information to answer the question.
c .The truck travels the same distance as the car.
d The truck travels half as far as the car.
Answer:
a)
Explanation:
Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:[tex]x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)[/tex]
Since the car starts from rest, v₀ =0.We know the value of t = 5 sec., but we need to find the value of a.Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:[tex]a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)[/tex]
Replacing a and t in (1):[tex]x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)[/tex]
Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:[tex]a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)[/tex]
Replacing v₀, at and t in (1), we have:[tex]x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)[/tex]
Therefore, as the truck travels twice as far as the car, the right answer is a).The effects of force on gravity is more noticeable when the object is_____
Answer:
Gravity only becomes noticeable when there is a really massive object like a moon, planet or star. We are pulled down towards the ground because of gravity. The gravitational force pulls in the direction towards the centre of any object.
An object with a mass of 8 kg moves at a speed of 5 m/s. How much kinetic energy does the object have?
Question 2 options:
100 J
200 J
40 J
20 J
Answer:
Kinetic energy =1/2mv^2
Where m is the mass of the object. V is the velocity
K.E=0.5(8*5^2)
K.E=100J
Explanation:
A satellite of mass m orbits a moon of mass M in uniform circular motion with a constant tangential speed of v. The satellite orbits at a distance R from the center of the moon. Write down the correct expression for the time T it takes the satellite to make one complete revolution around the moon?
The gravitational force exerted by the moon on the satellite is such that
F = G M m / R ² = m a → a = G M / R ²
where a is the satellite's centripetal acceleration, given by
a = v ² / R
The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that
v = 2πR / T → a = 4π ² R / T ²
Substitute this into the first equation and solve for T :
4π ² R / T ² = G M / R ²
4π ² R ³ = G M T ²
T ² = 4π ² R ³ / (G M )
T = √(4π ² R ³ / (G M ))
T = 2πR √(R / (G M ))
The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:
[tex] F = \frac{GmM}{R^{2}} [/tex] (1)
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
R: is the distance between the satellite and the center of the moon
m: is the satellite's mass
M: is the moon's mass
The gravitational force is also equal to the centripetal force:
[tex] F = ma_{c} [/tex] (2)
The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):
[tex] a_{c} = \frac{v^{2}}{R} [/tex] (3)
And from the tangential velocity we can find the period:
[tex] v = \omega R = \frac{2\pi R}{T} [/tex] (4)
Where:
ω: is the angular speed = 2π/T
By entering equations (4) and (3) into (2), we have:
[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex] (5)
By equating (5) and (1), we get:
[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]
[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]
[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]
[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]
Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
Find more here:
https://brainly.com/question/13340745?referrer=searchResultshttps://brainly.com/question/13451473?referrer=searchResultsI hope it helps you!
P and S waves from an earthquake travel at different speeds and this difference helps in locating the epicenter (point of origin) of the earthquake. (a) Assuming P waves travel at 10.3 km/s and S waves travel at 4.2 km/s, how far away did the earthquake occur if a particular seismic station detects the arrival of these two types of waves 3.25 minutes apart
Answer:
x = 1382.9 km
Explanation:
The speed of the wave is constant, so we can use the uniform motion relationships
p wave
[tex]v_p[/tex] = x / t₁
t₁ = x /v_p
S wave
v_s = x / t₂
t₂ = x / v_s
indicate that the time difference between the two waves is
t₂ - t₁ = 3.25 min (60 s / 1 min)
t₂ -t₁ = 195 s
let's substitute
[tex]\frac{x}{v_s} - \frac{x}{v_p}[/tex] = 195
x ([tex]\frac{1}{v_s} - \frac{1}{v_p}[/tex] = 195
let's calculate
x [tex]( \frac{1}{4.2} - \frac{1}{10.3} )[/tex] = 195
x (0.1410) = 195
x = 195 /0.141
x = 1382.9 km
Name the three types of kin etic energy. Define each.
Answer: Out of the 5, the first 3 that came to my mind first were thermal energy, electrical energy, and sound energy.
Explanation:
Thermal-Generates due to the quick motion of atoms and molecules, especially when they collide with each other. It is also called heat energy. The matters in the universe consist of atoms or molecules which are always in motion. However, we can’t see the movement of this energy with our naked eyes. We can feel it at the time it touches our skin.
Sound-The vibration of an object causes sound energy. Sound is the movement of energy generated by vibrations through some substance, such as air or water or solid. Sound energy can travel through any medium to transfer energy from one particle to another, and you can hear it. However, it cannot travel through a vacuum as a vacuum does not contain any particles that can act as a medium. When an object vibrates, it makes the surrounding particles vibrate by transferring its energy. These particles, when colliding with other particles, make them vibrate. In this way, the sound energy gets transferred from one particle to another.
Electrical- Every object in the universe is made up of small particles called atoms. Atoms consist of tiny particles, namely electrons, protons, and neutrons. The electrons in the atom always move around the nucleus of an atom. While applying voltage, the electrons present in the atom get energy and break the bonding with the parent atom and thus become free. The energy of this free electron is called electrical energy or electricity. Therefore, it is the energy of these moving free electrons. Electrons are negatively and positively charged and usually move through a wire.
If two clay masses produce a gravitational force of 340N, what will be the force if the distance is divided by 3 and the mass of one is divided by 2?
75.56 N
1530 N
510 N
56.67 N
Answer:
B: 1530 N
Explanation:
We know ghat formula for force of gravity is;
F_grav = G•m1•m2/d²
We are told that two clay masses produce a gravitational force of 340N.
Thus;
G•m1•m2/d² = 340
Now, if the distance is divided by 3 and the mass of one is divided by 2, we have;
F_g = (G × m1/2 × m2)/(d/3)²
Thus gives;
F_g = ½(G•m1•m2)/((1/9)d²)
Simplifying this gives;
F_g = (9/2)G•m1•m2/d²
From earlier, we saw that;
G•m1•m2/d² = 340
Thus;
F_g = (9/2) × 340
F_g = 1530 N
A bowling ball weighing 71.7 N is attached to the ceiling by a rope of length 3.73 m . The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.60 m/s. At this instant, what are:
a. the acceleration of the bowling ball, in magnitude and direction
b. the tension in the rope?
Answer:
A) a = 5.673 m/s²
The direction will be upwards vertically towards the point where it is suspended.
B) T = 113.2 N
Explanation:
A) We are given;
Weight of bowling ball; W = 71.7 N
Speed; v = 4.6 m/s
Rope length; r = 3.73 m
Now, formula for the centripetal acceleration is;
a = v²/r
Thus; a = 4.6²/3.73
a = 5.673 m/s²
The direction will be upwards vertically towards the point where it is suspended.
B) since weight is 71.7 N, it means that;
Mass = weight/acceleration = 71.7/9.8
Mass(m) = 7.316 kg
Thus,
Centripetal force is;
F_cent = 7.316 × 5.673
F_cent = 41.5 N
Thus, Tension in the rope is;
T = W + F_cent
T = 71.7 + 41.5
T = 113.2 N
A 0.140-kg baseball is pitched horizontally at 33.6 m/s. When a player hits the ball, it moves at the
same speed, but in the opposite direction. If the bat and the ball are in contact for 0.008 s, calculate the
average force the bat exerts on the ball.
Answer:
F = 1176 N
Explanation:
Given that,
Mass of a baseball, m = 0.140-kg
Initial speed of the baseball, u = 33.6 m/s
Final speed of the baseball, v = -33.6 m/s (in opposite direction)
The time of contact of the bat and the ball, t = 0.008 s
We need to find the average force the bat exerts on the ball. The force acting on the ball is given by :
[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.14(-33.6-33.6)}{0.008}\\F= -1176\ N[/tex]
So, the average force the bat exerts on the ball is 1176 N.
During the 1960's and 1970's, the Apollo spacecraft took humans to the moon in 3 days. Traveling to Mars requires a trip of about 2 astronomical units in total. How long would this trip take, traveling at the same speed as to the moon?
Answer:
join my z o o m
Explanation:
A bat, flying at 5.00 m/s, emits a chirp at 40.0 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat
Answer:
The answer is below
Explanation:
Firstly, the frequency is received by the wall and then it is reflected and received by the bat.
The frequency received by the wall (f') is given by:
f' = [tex]f(v\pm v_o)/v\\\\[/tex]
The - sign is used when the observer is moving away from the source and + sign when the observer is moving towards the source.
Since the bat is moving towards the wall, we use a positive sign. Hence:
f' = [tex]f(v+ v_o)/v\\\\[/tex]
The frequency reflected and received by the bat f" is:
f'' = [tex]f'\frac{v}{ (v\pm v_s)}\\\\[/tex]
- sign is used when the source moves toward the observer and + is used when the source moves away
since the bat moves towards the wall, then::
f'' = [tex]f'\frac{v}{(v-v_s)} =\frac{f(v+v_o)}{v}*\frac{v}{(v-v_s)} =f\frac{(v+v_o)}{(v-v_s)} \\\\[/tex]
v = speed of sound in air = 331 m/s, vo = velocity of observer = 5 m/s, vs = velocity of source = 5 m/s. Therefore:
[tex]f"=f\frac{(v+v_o)}{(v-v_s)} =40\ kHz\frac{(331\ m/s+5\ m/s)}{(331\ m/s+5\ m/s)} \\\\f"=41\ kHz[/tex]
The frequency of the echo received by the bat is 38.8 kHz.
The given parameters:
Speed of the bat, V = 5 m/sActual frequency of the chirp, Fs = 40 kHzSpeed of sound, Vs = 331 m/sThe observed frequency or frequency received by the bat is calculated by applying Doppler effect as follows;
[tex]f_0 = f_s(\frac{v \ +/-\ v_0}{v\ +/- \ v_s} )[/tex]
Since the bat is flying away from the wall the frequency received will be smaller than the actual frequency;
[tex]f_o = f_s (\frac{v \ - \ v_0}{v \ + \ v_s} )\\\\f_o = 40 \ kHz \times (\frac{331 - 5}{331 + 5} ) \\\\f_0 = 38.8 \ kHz[/tex]
Thus, the frequency of the echo received by the bat is 38.8 kHz.
Learn more about Doppler effect here: https://brainly.com/question/3841958
The skin temperature of a person is 34o C and his body surface area is about 1.8 m2 . He is standing bare skin in a room where the air temperature is 24o C and the walls are 17o C. He is metabolizing food at a rate of 155 W, the emissivity of his skin is 0.97 and there is a 5mm thick dead layer (immobile) air next to his skin acting as an insulation. a./ at what rate his body is losing heat by conduction
Answer:
the rate at which his body is losing heat by conduction is 93.6 J/s
Explanation:
Given that;
surface area A = 1.8 m²
Skin temperature of the person Tp = 32°C = ( 34 + 273.15 ) = 307.15 K
Temperature of Air [tex]T_{air}[/tex] = 24°C = ( 24 + 273.15 ) = 297.15 K
Temperature of wall [tex]T_{wall}[/tex] = 17°C = ( 17 + 273.15 ) = 290.15 K
Length ( thick dead layer = 5 mm = 0.005 m
Skin emissivity = 0.97
Rate of metabolism = 155 W
rate his body is losing heat by conduction = ?
first we determine the difference in temperature between the skin and air
so
ΔT = 307.15 K - 297.15 K = 10 K
we know that; coefficient of thermal heat conductivity of air k = 0.026 W/mK
so
rate of heat loss by conduction Q/ΔT will be;
Q/ΔT = (KA/L)ΔT
so we substitute
= ( 0.026 × 1.8/ 0.005 )10
= 9.36 × 10
= 93.6 J/s
Therefore, the rate at which his body is losing heat by conduction is 93.6 J/s
To wait until the oncoming vehicle passes before completing a left turn is known as:
a)
b)
c)
IPDE strategy
Risk acceptance
Risk rejection
Inappropriate maneuver
Answer:
Risk rejection
Explanation:
There are several factors that contribute to the degree of driving risks and they include but not limited to the ability of the driver and the condition of a vehicle. Other factors are condition of the environment and the condition of the highway. When driving, a driver may wait until an oncoming vehicle passes before making a complete left turn as a risk rejection strategy. Left turns are more dangerous when making them because drivers tend to accelerate on to a left turn. The wider radius of a left turn is know to led to higher speeds and greater pedestrian exposure. A driver is advised to have more mental and physical efforts when making a left turn.
The wave equation says that a waves __ is equal to its wavelength times is frequency.
Answer:
speed
Explanation:
The wave equation says that a waves speed is equal to its wavelength times is frequency.
What equal positive charges would have to be placed on two celestial objects with masses 4.69 × 1027 and 1.23 × 1025 kg to neutralize their gravitational attraction? (b) What mass of hydrogen would be needed to provide the positive charge calculated in (a)?
Answer:
q= 2.07*10¹⁶ C
m= 2.2*10⁸ kg.
Explanation:
a)
In order to neutralize their gravitational atraction (given by Newton's Universal Law of Gravitation), both masses must be repelled with an equal and opposite force, that must obey Coulomb's Law.So, we can write the following general equation:[tex]G*\frac{m_{1} *m_{2}}{r_{12}^{2}} = k * \frac{q_{1}*q_{2}}{r_{12} ^{2}} (1)[/tex]
Since G and k are universal constants, and that the distance on both sides is the same, and q₁ = q₂ = q, replacing in (1) the values of m₁ and m₂, we can solve for q, as follows:[tex]q =\sqrt{\frac{G*m_{1}*m_{2}}{k} } = \sqrt{\frac{6.67e-11*4.69e27*1.23e25}{9e9} } = 2.07e16 C (2)[/tex]
b)
Since hydrogen can carry only one elementary charge per atom, this charge will be distributed in a number of atoms that will be given by the charge q, divided by the value of the elementary charge, as follows:[tex]n_{H} = \frac{q}{e} =\frac{2.07e16C}{1.6e-19C} = 1.29e35 (3)[/tex]
Since each hydrogen has one proton and one electron, neglecting the mass of the electron, the mass of hydrogen needed to provide the positive charge q, will be just the product of the number of atoms needed (given by (3)) times the mass of the proton, as follows: [tex]m_{H} = n_{H} * m_{p} = 1.29e35*1.67e-27 kg = 2.2e8 kg (4)[/tex]a 75 kg object traveling at 4 m/s collides with and sticks to a 125 kg object initially at rest, what is the final velocity of the two objects?
Answer:
1.5m/s
Explanation:
Given parameters:
M1 = 75kg
V1 = 4m/s
M2 = 125kg
V2 = 0m/s
Unknown:
Final velocity of the two objects = ?
Solution:
To solve this problem, we must understand that this is an inelastic collision. To conserve momentum;
M1 V1 + M2 V2 = V (M1 + M2)
( 75 x 4 ) + ( 125 x 0) = V (75 + 125)
300 = 200V
V = 1.5m/s
Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground after
Select........seconds.
4.0
5.0
2.4
1.6
Answer and I will give you brainiliest
Answer:
2.4Explanation:
Hope it help mark as BrainlistA puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and puck of 0.167. If the puck is moving at an initial speed of 15.0 m/s, find the following.
(a) What is the force of kinetic friction? (Indicate the direction with the sign of your answer.)
N
(b) What is the acceleration of the puck? (Indicate the direction with the sign of your answer.)
m/s2
(c) How long does it take for the puck to come to rest?
s
(d) What distance does the puck travel during that time?
m
(e) What total work does friction do on the puck?
J
(f) What average power does friction generate in the puck during that time?
W
(g) What instantaneous power does friction generate in the puck when the velocity is 4.00 m/s?
W
Answer:
a) Ffr = -0.18 N
b) a= -1.64 m/s2
c) t = 9.2 s
d) x = 68.7 m.
e) W= -12.4 J
f) Pavg = -1.35 W
g) Pinst = -0.72 W
Explanation:
a)
While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.[tex]F_{frk} = -\mu_{k} * F_{n} (1)[/tex]
where μk is the kinetic friction coefficient, and Fn is the normal force.
Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:[tex]F_{n} = F_{g} = m*g (2)[/tex]
We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:[tex]F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)[/tex]
b)
According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.In this case, this net force is the friction force which we have already found in a).Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.We can write the expression for a as follows:[tex]a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)[/tex]
c)
Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:[tex]a = \frac{-v_{o} }{t} (5)[/tex]
Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:[tex]t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)[/tex]
d)
From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.So, we can use the following kinematic equation in order to find the displacement before coming to rest:[tex]v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)[/tex]
Since the puck comes to a stop, vf =0.Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:[tex]\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)[/tex]
e)
The total work done by the friction force on the object , can be obtained in several ways.One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:[tex]W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)[/tex]
f)
By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.[tex]P_{Avg} = \frac{\Delta E}{\Delta t} (10)[/tex]If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:[tex]P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)[/tex]
g)
The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:[tex]P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)[/tex]
please help fast due next period!!!
write the chemical formula for the following ionic compounds
Zinc (III) Phosphide
Answer:
The chemical formula for zinc (III) phosphide is Zn3P2
A point charge, Q1 = 12.0 C, is placed at the origin (0 cm, 0 cm) and a second charge, Q2, is placed at the coordinates (4.00 cm, 0 cm). A third charge, Q3 = 15.00 C, is placed at (5.0 cm, 0 cm). The force on Q3 is F⃑ = −20.0 N î. What is the value and sign of Q2?
Answer:
q₂ = -4.80 10⁻⁴ C = - 0.48 mC, charge is negative
Explanation:
Let's use coulomb's law
F = [tex]k \frac{q_1 q_2}{r^2}[/tex]
and the sum of forces, remember that charges of the same sign repel and of different sign attract
∑ F = F₁₃ + F₂₃ (1)
Let's start by fixing a reference system located at charge 1 with the positive direction to the right. In the problem it indicates that the net force on charge 3 is F = - 20.0 N, the negative sign indicates that the force is towards the left
let's look for every force, the charge q₁ = 12 10-⁻³ C and q₃ = 15 10⁻³ C
F₁₃ =[tex]k \frac{q_1 q_3}{x_{13}^2}[/tex]
F₁₃ = 9 10⁹ 12.0 15.0 10⁻⁶ / (5-0)²
F₁₃ = 64.8 10 3 N
This force is repulsive, that is, it is directed to the right
F₂₃ = k \frac{q_2 q_3}{x_{23}^2}
F₂₃ = 9 10⁹ q₂ 15.0 10⁻³ / (5-4)²
F₂₃ = 135 10⁶ q₂ N
we substitute in equation 1
-20.0 = 64.8 10³ + 135 10⁶ q₂
q₂ = (-20 - 64.8 10³) / 135 10⁶
q₂ = -4.80 10⁻⁴ C
the sign indicates that the charge is negative
A projectile leaves the muzzle of a projectile launcher with a speed of 35m/s [the
launcher is oriented vertically].
a) What is the maximum height reached by the projectile?
b) How long did it take the projectile to reach its maximum height?
c) How long was the projectile in the air [ttot]?
Answer:
a. 62.44 m
b. 3.57 sec
c. 7.12 sec
Explanation:
The leaving speed is given as = 35 m/s
The launcher is oriented vertically, thus the angle is = 0 degrees
a) The maximum height reached is given as;
h= v²₀y / 2g
h=35² / 2 * 9.81
h= 62.44 m
b) Time the projectile reaches maximum height is given as;
t=v₀y / g
t=35/9.81
t= 3.57 sec
c) The trip up is a mirror of the trip down
This means the time the projectile will be in the air is ;
t= 2 * 3.57
t= 7.12 sec
Astronaut 1 has a mass of 75 kg. Astronaut 2 has a mass of 80 kg. Astro 1 and 2 want to travel to separate planets, but they want to experience the same weight (in N). Astro 1 visits a planet with gravitational acceleration 12 m/s2. What must be Astro 2 planet's ag to equal Astro 1's weight
Answer:
weight = 900 N
acceleration = 11.25 m/s²
Explanation:
given data
mass m1 = 75 kg
mass m2 = 80 kg
gravitational acceleration = 12 m/s²
solution
As we know weight of a mass that is
weight of mass = Mass × Acceleration due to gravity .................1
so Astro 1 weight is
weight = 75kg × 12 m/s²
weight = 900 N
and
so, when Astro 2 needs this much weight the planet on which he is will have the acceleration
acceleration = Weight ÷ Mass of Astro 2 .....................2
acceleration = 900 ÷ 80 m/s²
acceleration = 11.25 m/s²
A baseball is thrown vertically into the air with a speed of 24.7 m/s.
How high does it go?
How long does the round trip up and down require?
PLEASE HEPPPP i need before midnight
I will do brainliest
Answer:
57.7
Explanation:
3.14 × 23 + - 78 = 57.7
A 50 kg crate slides down a 5.0 m loading ramp that is inclined at an angle of 25 to the horizontal. A worker pushes on the crate parallel to the surface of the ramp so that the crate slides down with a constant velocity. If the coefficient of kinetic friction between the crate and the ramp is 0.33, how much work is done by (a) the worker
Answer:
The magnitude of the work done by the worker is 303 J.
Explanation:
The work done by the worker can be found as follows:
[tex] W = |F|\cdot |d| cos(\alpha) [/tex]
Where:
F: is the force applied by the worker
d: is the displacement = 5.0 m
α: is the angle between the force applied and the displacement = 180°
We need to find the force applied by the worker:
[tex]\Sigma F = ma[/tex]
Taking as positive the movement direction of the crate we have:
[tex] -F - F_{\mu} + P_{x} = 0 [/tex]
Where:
m: is the crate's mass
a: is the acceleration = 0 (It is moving at constant speed)
F: is the force applied by the worker
Pₓ: is the weight in the horizontal direction
[tex]F_{\mu}[/tex]: is the frictional force
Hence, the force applied by the worker is:
[tex]F = P_{x} - F_{\mu} = mgsin(\theta) - \mu mgcos(\theta)[/tex]
[tex] F = 50 kg*9.81 m/s^{2}*(sin(25) - 0.33cos(25)) = 60.6 N [/tex]
Then, the work done by the worker is:
[tex] W = |F|\cdot |d| cos(\alpha) = 60.6 N*5.0 m*cos(180) = -303 J [/tex]
Therefore, the magnitude of the work done by the worker is 303 J.
I hope it helps you!
The work is done by the worker will be 303 J. Work done is described as the multiplication of applied force and the amount of displacement.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
F is the force applied by the worker
d is the displacement = 5.0 m
α is the angle between the force applied and the displacement = 180°
m is the crate's mass= 50 kg
a is the acceleration = 0
Pₓ: is the weight in the horizontal direction
The net force on the crate is found as;
[tex]\rm F_{net}= P_X - F_{\mu} \\\\ \rm F_{net}= mg sin \theta - \mu mg cos \theta \\\\ \rm F_{net}=50 \times \times 9.81 (sin 25^0 -0.33 cos(25) \\\\ \rm F_{net}= 60.6 N \\\\[/tex]
The work done by the worker will be;
[tex]\rm W= Fd cos \alpha \\\\ \rm W= 60.6 \times 5.0 cos 180^0 \\\\\rm W=-303 \ J[/tex]
Hence the work is done by the worker will be 303 J.
To learn more about the work done refer to the link ;
https://brainly.com/question/3902440
The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?
Answer:
The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.
Explanation:
Statement is incomplete. Complete description is presented below:
A freight train has a mass of [tex]1.83\times 10^{7}\,kg[/tex]. The wheels of the locomotive push back on the tracks with a constant net force of [tex]7.50\times 10^{5}\,N[/tex], so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?
If locomotive have a constant net force ([tex]F[/tex]), measured in newtons, then acceleration ([tex]a[/tex]), measured in meters per square second, must be constant and can be found by the following expression:
[tex]a = \frac{F}{m}[/tex] (1)
Where [tex]m[/tex] is the mass of the freight train, measured in kilograms.
If we know that [tex]F = 7.50\times 10^{5}\,N[/tex] and [tex]m = 1.83\times 10^{7}\,kg[/tex], then the acceleration experimented by the train is:
[tex]a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}[/tex]
[tex]a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}[/tex]
Now, the time taken to accelerate the freight train from rest ([tex]t[/tex]), measured in seconds, is determined by the following formula:
[tex]t = \frac{v-v_{o}}{a}[/tex] (2)
Where:
[tex]v[/tex] - Final speed of the train, measured in meters per second.
[tex]v_{o}[/tex] - Initial speed of the train, measured in meters per second.
If we know that [tex]a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex] and [tex]v = 22.222\,\frac{m}{s}[/tex], the time taken by the freight train is:
[tex]t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }[/tex]
[tex]t = 542.265\,s[/tex]
The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.
