Answer:
33.5J
Explanation:
Given:
Mass of the canister= 3.1 kg
Initial velocity of canister= v(i)= 4.4i m/s
Final velocity of canister= v(f)= 6.4j m/s
Force magnitude( xy plane)= 5 N
The magnitude of vector V'= Vxi + Vyj + Vzk
|V|= √( Vx^2 + Vy^2 + Vz^2
From Kinectic energy and work theorem.
Net work = Kinectic energy of the canister
ΔK= W
(Kf - Ki)= W
Where Kf= final Kinectic energy
= 1/2 mv^2
If we input the given values we have,
= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J
Ki= initial Kinectic energy
= 1/2 mv^2
If we substitute the given values we have
=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J
Work done by canister = (final Kinectic energy - initial energy)
= 63.5- 30
=33.5J
Hence, work done on the canister 33.5J
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.
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
how much work is required to make a 1400 kg car increase its speed from 10 m/s to 20 m/s?
what average force is required if the car travels 15 m during this speed change?
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
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
You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of electrons from each sphere.
1) What is the charge on each sphere if their gravitational attraction is exactly equal to their electrical repulsion?
2) How many electrons did you remove from each sphere?
Answer:
q = 2.066* 10⁻¹³ C.
n = 1,291,250 electrons.
Explanation:
1)
If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:[tex]F_{g} = F_{c} (1)[/tex]
where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:[tex]F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)[/tex]
Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:[tex]F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)[/tex]
since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:[tex]G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)[/tex]
Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:[tex]Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)[/tex]
Since both charges are the same, the charge on each sphere is just the square root of (5):Q = 2.066* 10⁻¹³ C.2)
Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:[tex]n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)[/tex]What does Newtons 1st Law Desribe?
A. the tendency of moving objects to stay in motion
B. How force balances out of zero net force
C. The change in an objects motion over time
D. The attractive force between two objects due to their masses
Answer:
A. the tendency of moving objects to stay in motion
A quarterback, Patrick, throws a football down the field in a long arching trajectory to wide receiver, Tyreek. The football and Tyreek are traveling in the same direction, started at the same spot, and the football was thrown at the same instant that Tyreek began running. Furthermore, both Tyreek and the football have horizontal components of speed of 22.6 mph. Under these circumstances, no matter what angle the football is thrown at, it will land on Tyreek (whether he catches it or not). True or false
Answer:
the statement is False [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
Explanation:
Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.
Let's start by finding the time it takes for the ball to reach the ground
y = [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground its height is y = 0
0 = vo sin θ - ½ g t²
0 = t (vo sin θ - ½ g t)
The results are
t = 0 exit point
t = 2 v₀ sin θ/g arrival point
at this point the ball traveled
[tex]x_{ball}[/tex]= v₀ₓ t
x_{ball} = v₀ cos θ 2v₀ sin θ / g
x_{ball}= 2 v₀² cos θ sin θ/ g
Now let's find that distantica traveled the receiver in time
[tex]x_{rec}[/tex] = v₀ t
x_{rec} = v₀ (2 v₀ sin θ / g)
x_{rec} = 2 v₀² sin θ / g
without dividing this into two distances
[tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
therefore the distances are not equal to the ball as long as behind the receiver
Therefore the statement is False
A student in gym class swings from a rope and they are moving 5 m/s at the bottom of their swing. What is the height they reach above the floor before swinging back down?
1/2v^2=gh g = 9.8 m/s^2
A 2.55 m
B. 1.28 m
c. 5m
D. 12.5 m
Answer:
A
Explanation:
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
Please help me I’ll mark brainless .
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
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
There is always one way to calculate the magnitude of the net vector of any direction vectors.
True
False
Answer and I will give you brainiliest
Answer:
False
Explanation:
Hope it help mark as BrainlistAnswer:
False
Explanation:
Suppose a car is traveling at 18.6 m/s, and the driver sees a traffic light turn red. After 0.500 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 4.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light
Answer:
s = 52.545 m
Explanation:
First, we calculate the distance covered during the 0.5 s when the driver notices the light and applies the brake.
[tex]s_{1} = vt\\[/tex]
where,
s₁ = distance covered between noticing light and applying brake = ?
v = speed = 18.6 m/s
t = time = 0.5 s
Therefore,
[tex]s_{1} = (18.6\ m/s)(0.5\ s)\\s_{1} = 9.3\ m\\[/tex]
Now, we calculate the distance for the car to stop after the application of brakes. For that we use 3rd equation of motion:
[tex]2as_{2} = V_{f}^{2} - V_{i}^{2}\\\\[/tex]
where,
s₂ = distance covered after applying brake = ?
a = deceleration = - 4 m/s²
Vf = final speed = 0 m/s
Vi = initial speed = 18.6 m/s
Therefore,
[tex]2(- 4\ m/s^{2})s_{2} = (0\ m/s)^{2} - (18.6\ m/s)^{2}\\\\s_{2} = \frac{(18.6\ m/s)^{2})}{8\ m/s^{2}}\\\\s_{2} = 43.245\ m[/tex]
So the total distance covered by the car before stopping is:
[tex]s = s_{1} + s_{2}\\s = 9.3\ m + 43.245\ m\\[/tex]
s = 52.545 m
Need help in the question that is down
Answer:
4th ans
Explanation:
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 electron moving along the x axis has a position given by x 16te$t m, where t is in seconds. How far is the electron from the origin when it momentarily stops
This question is incomplete, the complete question is;
An electron moving along the x axis has a position given by x = 16 t[tex]e^{-t}[/tex] m, where t is in seconds. How far is the electron from the origin when it momentarily stops
Answer:
the electron is 5.88 m far from the origin when it momentarily stops
Explanation:
Given that;
the position of the electron is x = 16 t[tex]e^{-t}[/tex] m
now, if the electron stopped after a time t, then its velocity is zero
so
V = dx/dt = 0
d/dt( 16 t[tex]e^{-t}[/tex] m) = 0
16( -t[tex]e^{-t}[/tex] + [tex]e^{-t}[/tex] ) = 0
16(-t + 1) [tex]e^{-t}[/tex] = 0
16(1 - t) [tex]e^{-t}[/tex] = 0
1 - t = 0
t = 1 sec
so
x = 16 × 1 × [tex]e^{-1}[/tex] m
x = 16 × 1 × 0.36787 m
x = 5.88 m
Therefore, the electron is 5.88 m far from the origin when it momentarily stops
Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose, instead, that the Boxster is initially 170 m behind the Scion. The speed of the Boxster is 24.4 m/s and the speed of the Scion is 18.6 m/s. How much time does it take for the Boxster to catch the Scion
Answer:
It will take 29.31 seconds for the Boxster to catch the Scion
Explanation:
Given the data in the question;
lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A
the distance travelled by car A is
x = [tex]V_{A}[/tex] × t
where [tex]V_{A}[/tex] is the speed of the car and t is time
the distance travelled by car B before reaching car A will be;
x + x₀ = [tex]V_{B}[/tex] × t
Now lets replace x by [tex]V_{A}[/tex] × t
so
([tex]V_{A}[/tex] × t) + x₀ = [tex]V_{B}[/tex] × t
x₀ = ([tex]V_{B}[/tex] × t) - ([tex]V_{A}[/tex] × t)
x₀ = t ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])
t = x₀ / ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])
so we substitute
t = 170 m / (24.4 - 18.6)
t = 170 / 5.8
t = 29.31 s
Therefore; it will take 29.31 s for the Boxster to catch the Scion
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:
i will give the brainliest answer to whoever answers this, explain how Sir William Gilbert used models in his investigation of magnetism
Explanation:
Lodestone am iron were the only known magnetic materials in Gilbert's day, and his task was to investigate magnetism. Gilbert was so sure that the earth was a giant lodestone and used the earth as a primary reference, defining the north(magnetic) pole of a needle, or a a nail floating on a piece of cork, to be that which turns towards the Earth's north geographic pole. he wanted to prove this with a model Terella, using short pieces of iron.
Magnetic dipole X is fixed and magnetic dipole Y is free to move. Dipole Y will initially:
A. Move toward X but not rotate
B. Move away from X but not rotate
C. Move toward X and rotate Move away from X and rotate
D. Rotate but not translate
Answer: The correct option is A.
Move toward X but not rotate
Explanation:
This is because from the question, X is fixed and y is free to move. Since magnetic dipole of X is fixed, that is it can't move and that of y is free to move, therefore y will move toward x because it's forces of attraction is linear and not rotational and besides X and Y are on a linear path, therefore Y will move towards X that is fixed and it will not rotate since it's linear.
A boy standing at one end of a floating raft that is stationary relative to the shore walks to the opposite end of the raft, away from the shore. As a consequence, the raft does which of the following?
a. remains stationary
b. moves away from the shore
c. moves toward the shore.
Answer:
The raft moves towards the shore
Explanation:
Two skaters, each of mass 45 kg, approach each other along parallel paths separated by 3.2 m. They have equal and opposite velocities of 1.1 m/s. The first skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes.. Assume frictionless ice. Describe quantitatively the motion of the skaters after they have become connected by the pole. What is their angular speed?
Answer:
Explanation:
The skaters in a circle of radius; R = L/2
R = 3.2/2 = 1.6 m
From Ii*Wi = If*Wf
Note; i is initial while f is final
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
What name is given to "an object or group of objects that we wish to consider for a physics problem?
Answer:
System
Explanation:
A system is an object or group of objects that we wish to consider for a physics problem. Examples of systems are universe, house and its surroundings etc.
There are three different types of systems which are:
Isolated system: In this type of system there is no energy or matter exchange with the surroundings. Closed system: In this type of system, only energy is exchanged with the surroundings, matter is not exchanged. Open system: in this system both matter and energy are exchanged with the surroundings..The chain of DNA consists of
ILL MARK BRAINLIST !! get it right
In which of Earth’s systems would we find cloud droplets, wind, & weather?
Cryosphere
Geosphere
Biosphere
Atmosphere
Answer:
atmosphere
Explanation:
.........,....
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
After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38 m/s on flat ground. Assuming that his car travels with a constant acceleration, calculate the maximum speed (Vm), in meters per second, his car can reach on flat ground?\
Answer:
64.2 m/s
Explanation:
We are given that
Speed ,v=38 m/s
We have to find the maximum speed when his car reach on flat ground.
Using dimensional analysis
[tex]F_{res}\propto v^2[/tex]
If 35% acceleration reduced by F(res) at 38 m/s
Then, 100% acceleration can be reduced by F(res) at v' m/s
[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]
[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]
[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]
Substitute the values
[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]
[tex]v'=64.2 m/s[/tex]
Hence, the maximum speed when his car can reach on flat ground=64.2 m/s
The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
What is the relation between resistance and speed?
The air resistance is directly proportional to the square of the velocity of an object.
R ∝ v²
The speed of the car was reduced by 35 % at 38 m/s.
So, the speed of the car was reduced by 100% at v' m/s.
The relationship can be given by,
[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]
Put the values in the formula and calculate for [tex]v'[/tex],
[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]
[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]
Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
Learn more about resistance and speed?
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