Given that the charge of the toaster is q =700 C
The current of the toaster is I = 10 A
We have to find the time and number of electrons.
Time can be calculated by the formula,
[tex]t=\frac{q}{I}[/tex]Substituting the values, the time will be
[tex]\begin{gathered} t=\frac{700}{10} \\ =70\text{ s} \end{gathered}[/tex]The number of electrons can be calculated by the formula,
[tex]n=\frac{q}{e}[/tex]Here, n is the number of electrons
and e is the charge of the electron whose value is
[tex]1.6\text{ }\times10^{-19}\text{ C}[/tex]Substituting the values, the number of electrons will be
[tex]\begin{gathered} n\text{ = }\frac{700}{1.6\times10^{-19}} \\ =4.375\text{ }\times10^{-17} \end{gathered}[/tex]A 20.0 kg penguin slides at a constant velocity of 3.3 m/s down an icy incline. The incline slopes above the horizontal at an angle of 6.0°. Determine the coefficient of kinetic friction.
The coefficient of kinetic friction down the slope is 0.105.
What is kinetic friction?Kinetic friction is the friction that exists or acts between the surfaces of one object moving over another.
The kinetic frictional force of an object moving on an inclined plane is give by the formula below:
Kf = μk * mg *cosθ
where;
μk = coefficient of kinetic friction.
mg cosθ = component of the weight perpendicular to the inclined plane
θ = angle of inclination
For an object moving at a constant velocity, the component of the weight down the slope (mg sinθ) is equal to the kinetic frictional force.
Hence, μk * mg *cosθ = mg sinθ
μk = mg sinθ / mg *cosθ
μk = tan θ
μk = tan 6.0
μk = 0.105
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The total mechanical energy of the roller coaster cart below at Point A is 180,000 J. The speed of the cart at Point B is +20 m/s. Assume no energy is lost due to dissipative forces such as friction. A) What is the mass (in kg) of the roller coaster cart? B) What is the potential energy at Point A? C) What is the kinetic energy at Point A?
Mechanical energy (ME) = Potential energy(PE) + kinetic energy (KE)
PE = mgh
m= mass
g= gravity
h= height
KE= 1/2 m v^2
v= speed
Point B
ME = KE + PE
PE = 0 (height = 0 )
KE = 1/2 (m) v^2
180,000 = 1/2 (m) (20)^2
m = 180,000 / (1/2 (20)^2 )
m= 900 kg
Point A.
ME = 180,000 J = PEa + KE a
PEa = m g h = 900 (9.8) (20) = 176,400J
MEa = PEa + KEa
KEa = MEa - PEa = 180,000 - 176,400 J = 3,600 J
A) mass = 900 kg
B) 176,400 J
C) 3,600 J
A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?
Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:
[tex]\begin{gathered} d=vt \\ =(3.1\frac{m}{s})(314s) \\ =973.4m \end{gathered}[/tex]That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:
[tex]\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=\frac{d}{2\pi} \\ =\frac{973.4m}{2(3.14...)} \\ =154.9...m \end{gathered}[/tex]The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:
[tex]a_c=\frac{v^2}{r}[/tex]Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:
[tex]a_c=\frac{(3.1\frac{m}{s})^2}{(154.9m)}=0.062\frac{m}{s^2}[/tex]Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.
8. What would the shape of the graph be if the speed of the object decreased from 50.0 km/hr at 20 s to 30 km/hr at 40 s?9. What is the acceleration in Problem 8?
We will have the following:
8. The graph will be the following:
9. The acceleration of part 8 will be determined as follows:
First, we determine the equivalency of the velocities, that is
[tex]\begin{gathered} \frac{50km}{h}\ast\frac{1000m}{1km}\ast\frac{1h}{3600s}=\frac{125}{9}m/s \\ \\ \frac{30km}{h}\ast\frac{1000m}{1km}\ast\frac{1h}{3600s}=\frac{25}{3}m/s \end{gathered}[/tex]Then, the acceleration will be:
[tex]a=\frac{(25/3\text{ }m/s)-(125/9\text{ }m/s)}{40s-20s}\Rightarrow a=-\frac{5}{18}m/s^2[/tex]So, the acceleration was of -5/18 m/s^2.
Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net. With what speed did he leave the floor?
Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net, the speed from which he would have left the floor would be 4.64 m / s .
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net.
By using the third equation of the motion,
v² - u² = 2 × a × s
0 - u² = 2 × -9.81 × 1.10
u = 4.64 m / s
Thus, the speed from which he would have left the floor would be 4.64 m / s .
To learn more about equations of motion here, refer to the link given below;
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I want to know if I’m doing this correctly, and if not, then what would be the best process to do solve it?
Given
[tex]3\times10^{-10}m^{}[/tex]
Factor Name Symbol
10-1 deci d
10-2 centi c
10-3 milli m
10-6 micro µ
10-9 nano n
10-12 pico p
10-15 femto f
10-18 atto a
10-21 zepto z
10-24 yocto y
a. nanometers
0.3 nm
[tex]3\times10^{-1}nm^{}[/tex]b. picometers
300 pm
[tex]3\times10^2pm[/tex]What is the initial velocity of an automobile acquiring a final velocity of 32 m/s with an acceleration of 4.0m/s ²
Answer:
Explanation:
Given:
V = 32 m/s
a = 4.0 m/s²
__________
V₀ - ?
V = V₀ + a*t
V₀ = V - a*t = 32 - 4*t
Time is not set according to the condition of the problem!
There's not enough given information t o answer the question. It depends on how long the car has been accelerating.
it could be 28 m/s 1 second ago.
it could be 16 m/s 4 seconds ago.
it could be 10 m/s 5.5 seconds ago.
etc.
i'll take a wild guess and speculate that the question actually tells how long the car has been accelerating, but you didn't copy that part.
A blink of an eye is a time interval of about 150ms for an average adult. The closure portion of the blink takes only about 55ms. Let us model the closure of the upper eyelid as uniform angular acceleration through an angular displacement of 16.6 degree. What is the value of the angular acceleration the eyelid undergoes while closing 2. What is the tangential acceleration of the edge of the eyelid while closing if the radius of the eyeball is 1.25 cm?
ANSWER:
STEP-BY-STEP EXPLANATION:
The first thing is to convert the time into a second, just like this:
[tex]t=55\text{ ms}\cdot\frac{1\text{ s}}{1000\text{ ms}}=0.055\text{ s}[/tex]Now, convert the angular displacement of the eyelid from degrees to rad:
[tex]\partial\theta=16.6\text{\degree}\cdot\frac{2\pi\text{ rad}}{360\text{\degree}}=0.29\text{ rad}[/tex]We can calculate the angular velocity, dividing the angular momentum by the time, like this:
[tex]w=\frac{0.29}{0.055}=5.27\text{ rad/s}[/tex]The angular acceleration is calculated by means of the quotient of the difference in angular velocity and time, like this:
[tex]a_w=\frac{\delta w}{\delta t}=\frac{5.27-0}{0.15-0.055}=55.47\text{ rad/s}^2[/tex]the tangential acceleration would be:
a) your one friend and their bumper boat (m = 210 kg) are traveling to the left at 3.5 m/s and your other friend and their bumper boat (m = 221 kg) are traveling in the opposite direction at 1.8 m/s. The two boats then collide in a perfectly elastic one-dimensional collision. How fast is your first friend and their boat moving after the collision?b) after the collision, a constant drag force between the water and the boat causes your first friends boat to come to a stop. If the boat travels 2.7 m before stopping, what is the magnitude of the constant drag force?
ANSWER:
a) 1.757 m/s
b) 119.91 N
STEP-BY-STEP EXPLANATION:
Given:
Mass 1 (m1) = 210 kg
Initial speed 1 (u1) = 3.5 m/s
Mass 2 (m2) = 221 kg
Initial speed (u2) = 1.8 m/s
We make a sketch of the situation:
a)
We make a momentum balance by taking into account the conservation of momentum:
[tex]\begin{gathered} m_1u_1-m_2u_2=m_1v_1+m_2v_2 \\ \\ v_2=\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\rightarrow(1) \end{gathered}[/tex]Now an energy balance taking into account the conservation of energy, as follows:
[tex]\begin{gathered} \frac{1}{2}m_1(u_1)^2+\frac{1}{2}m_2(u_2)^2=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\lparen v_2)^2\rightarrow(2) \end{gathered}[/tex]Now, we substitute equation (1) in (2) and we get the following:
[tex]\begin{gathered} m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\right)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{(m_2)^2}\right) \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{m_2}\right) \\ \\ v_1=u_1\frac{m_1-m_2}{m_1+m_2}+u_2\frac{2m_2}{m_1+m_2} \\ \\ \text{ Now, we substitute each value, like so:} \\ \\ v_1=3.5\cdot\frac{210-221}{210+221}+1.8\cdot\frac{2\cdot221}{210+221} \\ \\ v_1=1.757\text{ m/s} \end{gathered}[/tex]b)
We use the following formula to determine the force:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ Wee replacing:} \\ \\ (1.756)^2=0^2+2\cdot a\cdot2.7 \\ \\ a=0.003375\text{ m/s}^2 \\ \\ \text{ Therfore:} \\ \\ F=m\cdot a=210\cdot0.571=119.91\text{ N} \end{gathered}[/tex]Which resistors in the circuit must always have the same current?A.B and CB.A and BC.C and DD.A and D
ANSWER
D. A and D
EXPLANATION
Two resistors have the same current if they are connected in series. As we can see in the schematic, resistors B and C are connected in parallel, so they don't have the same current - unless they have the same resistance.
Resistors A, D, and the equivalent of the parallel resistors (B and C) are connected in series, so they always have the same current.
Hence, of these options, resistors A and D always have the same current.
how many joules does a lamp marked 12 volts, 24 w consumed in an hour . And also what is the current?
Given data:
* The voltage given is 12 volts.
* The value of the power given is 24 Watt.
* The time till which the lamp is used is,
[tex]\begin{gathered} t=1\text{ hr} \\ t=60\times60\text{ s} \\ t=3600\text{ s} \end{gathered}[/tex]Solution:
(a). The energy consumed by the lamp in time t is,
[tex]E=P\times t[/tex]where P is the power of lamp, t is the time, and E is the energy consumed,
Substituting the known values,
[tex]\begin{gathered} E=24\times3600 \\ E=86400\text{ J} \\ E=86.4\times10^3\text{ J} \\ E=86.4\text{ kJ} \end{gathered}[/tex]Thus, the energy consumed by the lamp in 1 hour is 86.4 kJ.
(b). The power of the lamp in terms of voltage and current is,
[tex]\begin{gathered} P=VI \\ I=\frac{P}{V} \end{gathered}[/tex]where P is the power, V is the voltage and I is the current,
Substituting the known values,
[tex]\begin{gathered} I=\frac{24}{12} \\ I=2\text{ A} \end{gathered}[/tex]Thus, the current flowing through the lamp is 2 A.
You place a box weighing 276 N on an inclined plane that makes a 44.5° angle with the horizontal.
Compute the component of the gravitational force acting down the inclined plane.
Answer in units of N.
Answer:
this is the answer
Explanation:
hope it helps
A 2.0 microF capacitor is connected across a 60 Hz voltage source, and a current of 2.0 mA is measured on a VOM. What is the capacitive reactance of the circuit?
Let's write down and name the variables we know.
C: capacitance; C = 2 μF = 2*10^-6 F
f: frequency of voltage source; f = 60 Hz
Xc: capacitive reactance of circuit (we are solving for this)
We also know that ω = 2πf = 120π.
From this information, we can use the following equation:
Xc = 1/(ωC)
And we can solve for Xc.
Xc = 1/(120π*2*10^-6)
Xc = 1326.291 Ω
As a torque activity, your Physics TA sets up the arrangement decribed below. A uniform rod of mass mr = 158 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the r1 = 10.0 cm and r2 = 90.0 cm mark, passed over pulleys, and masses of m1 = 281 g and m2 = 177 g are attached. Your TA asks you to determine the following: (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). r3 = Fp = F = (b) Let's now remove the mass m3 and determine the new mass m4 you would need to suspend from the rod at the position r4 = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 = Fp = F = (c) Let's now remove the mass m4 and determine the mass m5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 = r5 =
a) Recall, the net torque on the rod must be zero. Thus,
Σt = 0
where
t represents torque
Thus,
t1 + t2 - tr - t3 = 0
t = rF
where
F = force
r = distance
r1F1 + r2F2 - rrFr - r3F3 = 0
r3F3 = r1F1 + r2F2 - rrFr
r3 = (r1F1 + r2F2 - rrFr)/F3
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
Thus,
r3 = (r1m1g + r2m2g - rrmrg)/m3g
g cancels out
r3 = (r1m1 + r2m2 - rrmr)/m3
From the information given,
r1 = 10 cm = 10/100 = 0.1 m
r2 = 90 cm = 90/100 = 0.9 m
rr = 100/2 = 50 cm = 50/100 = 0.5 m
m1 = 281 g = 281/1000 = 0.281 kg
m2 = 177g = 0.177 kg
mr = 158g = 0.158 kg
m3 = 200g = 0.2kg
By substituting these values into the equation,
r3 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
r3 = 0.542 m
The force exerted by the pin, Fp = mg
g = 9.8
Fp = (m3 - mr - m1 - m2)g
Fp = (0.2 + 0.158 - 0.281 - 0.177)9.8
Fp = - 0.981
Taking the absolute value,
IFpI = 0.981 N
F = - 90 degrees
b) r1F1 + r2F2 - rrFr - r4F4 = 0
r4F4 = r1F1 + r2F2 - rrFr = 0
F4 = (r1F1 + r2F2 - rrFr)/r4
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
F4 = T4 = m4g
Thus,
m4g = (r1m1g + r2m2g - rrmrg)/r4
m4g = (r1m1 + r2m2 - rrmr)/r4
r4 = 0.2
By substituting these values into the equation,
m4 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
m4 = 0.542 kg
The force exerted by pin is
Fp = (m4 + mr - m1 - m2(g
Fp = (0.542 + 0.158 - 0.281 - 0.177)9.8
Fp = 2.37 N
Fp = 2.37 N
F = 90 degrees
c) When the pin does not exert a force,
Fp = 0
F1 + F2 - Fr = F5
m1 + m2 - mr = m5
m5 = 0.281 + 0.177 - 0.158
m5 = 0.3 kg
Since the net torque on the rod is zero,
t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
r5 = r1F1 + r2F2 - ffFr)/F5
r5 = (r1m1 + r2m2 - rrmr)/m5
r5 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.3
r5 = 0.36
If a space ship traveling at a 1000 miles per hour enters and area free of Gravitational forces,it’s engine must run at some minimum level in order to maintain the ships velocity. is this statement true or false
The given statement is false.
If a spaceship traveling at 1,000 miles per hour enters an area free of gravitational forces, its engine must run at some maximum level in order to maintain the ship’s velocity
What do you diagram to analyze orbital motion ?
The diagram to analyze the orbital motion can be shown as,
Here, a is the acceleration of moon and v is the speed.
The above diagram indicates the orbital motion of the moon around the earth. The moon is more towards the earth than the sun due to larger gravity of earth and at the same time the moon has its velocity that tends moon to move. Therefore, the moon has balanced gravitational and centripetal force to keep in an uniform orbital motion.
order the colors of the discuses to show the size of the force applied to throw each discus
Since the acceleration will be the same for all, the order would be the following [From larger to smaller force]:
Blue
Green
Orange
Red
Purple
Need help 82x2682 please
ANSWER:
STEP-BY-STEP EXPLANATION:
We have the following multiplication:
[tex]undefined[/tex]Given the wave described by y(x,t)=5cos[π(4x-3t)], in meters. Find the following. Giveexact answers with units.
Answer:
a) 5 m
b) 0.667 s
c) 0.5 m
d) 0.75 m/s
e) -5 m
Explanation:
In an equation of the form
y(x, t) = Acos(kx - ωt)
A is the amplitude, ω = 2π/T where T is the period, and k = 2π/λ where λ is the wavelength. In this case, the equation os
y(x,t) = 5cos(π(4x - 3t)
y(x,t) = 5cos(4πx - 3πt)
So, A = 5, k = 4π, and ω = 3π. Then, we can find each part as follows
a) Amplitude
The amplitude is A, so it is 5 m.
b) the period
Using the equation ω = 2π/T and solving for T, we get:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{3\pi}=\frac{2}{3}=0.667\text{ s}[/tex]So, the period is 0.667 s
c) the wavelength.
using the equation k = 2π/λ and solving for λ, we get:
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{4\pi}=0.5\text{ m}[/tex]So, the wavelength is 0.5 m
d) The wave speed
The wave speed can be calculated as the division of the wavelength by the period, so
[tex]v=\frac{\lambda}{T}=\frac{0.5\text{ m}}{0.667\text{ s}}=0.75\text{ m/s}[/tex]e) The height of the wave at (2, 1)
To find the height, we need to replace (x, t) = (2, 1) on the initial equation, so
[tex]\begin{gathered} y(x,t)=5\cos(\pi(4x-3t)) \\ y(2,1)=5\cos(\pi(4\cdot2-3\cdot1)) \\ y(2,1)=5\cos(\pi(8-3)) \\ y(2,1)=5\cos(\pi(5)) \\ y(2,1)=5\cos(5\pi) \\ y(2,1)=5(-1) \\ y(2,1)=-5 \end{gathered}[/tex]Then, the height of the wave is -5 m.
Therefore, the answers are
a) 5 m
b) 0.667 s
c) 0.5 m
d) 0.75 m/s
e) -5 m
calculate how much heat is released when 500g of platinum is cooled from 250.0K to 240.0K
Answer:
Q = 665 J
Explanation:
Given:
m = 500 g = 0.5 kg
T₁ = 250.0 K
T₂ = 240.0 K
c = 133 J / ( kg*⁰K) - Specific heat capacity of platinum
___________
Q - ?
Heat is released:
Q = c*m*( T₁ - T₂) = 133*0.5*(250.0 - 240.0) =
= 133*0.5*10 ≈ 665 J
some more disconnectiob
let the student report the tutor and check
answer is updated
answer is updated olalalala
olalalala 2
olalalala 3
The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley.
Answer: It should be the answer beginning like this
The linear spread is that
The radial acceleration of locations along the blade's outer edge is approximately 17580 [tex]m/s^2[/tex].
What is Radial acceleration?Radial acceleration describes the acceleration of an object travelling on a circular path towards the circle's center. It can be defined as the rate of change of tangential velocity with regard to time and is also known as centripetal acceleration.
Given:
A table saw's engine rotates at 3450 revolutions per minute.A V-belt connects a pulley that's attached to the motor shaft to a second pulley half the diameter.A 0.208 m circular saw blade is installed on the same rotating shaft as the second pulley.We know that the motor is rotating at 3450 rev/min. One revolution is equal to 2π radians, so we can convert the motor speed to radians per minute:
ω₁ = (3450 rev/min) x (2π rad/rev) = 21675π rad/min
The second pulley is half the diameter of the first pulley, so its angular speed, ω₂, is twice that of ω₁:
ω₂ = 2ω₁ = 43350π rad/min
The circular saw blade is mounted on the same shaft as the second pulley, so it also rotates at the same angular speed:
ω = ω₂ = 43350π rad/min
We can now calculate the linear speed of the small piece of wood moving at the same rate as the rim of the circular saw blade, indicated by v. The circumference of the circle is supplied by the rim of the circular saw blade:
C = πd = π(0.208 m) = 0.6548 m
The linear speed of the little piece of wood is equal to the tangential speed of the circular saw blade's rim:
v = ωr
where r is the circular saw blade's radius, given by half its diameter:
r = d/2 = 0.208/2 = 0.104 m
By substituting the values, we obtain:
v = r = (43350 rad/min) x (0.104 m) x (1/60) = approx. 23.0 m/s
As a result, the linear speed of the little piece of wood moving at the same rate as the rim of the circular saw blade is about 23.0 m/s.
Next, compute the radial acceleration of locations on the blade's outer edge, represented by. The radial acceleration is calculated as follows:
α = rω²
By substituting the values, we obtain:
r2 = (0.104 m) x (43350 rad/min)2 x (1/602) = 17580 m/s2 (approximate)
Therefore, the radial acceleration of points on the outer edge of the blade is approximately 17580 m/s². This high radial acceleration explains why sawdust doesn't stick to the teeth of the saw blade.
Learn more about Radial acceleration, here:
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Your question is incomplete, most probably the complete question is:
The motor of a table saw is rotating at 3450{\rm rev/min}. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208{\rm m}is mounted on the same rotating shaft as the second pulley.
The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed?
v =_______________________ m/s
Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.
\alpha=______________________m/s2
A sound wave of wavelength 1.66m at a temperature of 23 C is produced for 2.5 seconds. How far does this wave travel? How many complete waves are emitted in this time interval?
To find how far it travels, we just have to multiply
[tex]1.66\times2.5=4.15[/tex]It travels 4.15 meters.
During this interval, there are emitted 2.5 waves.
Three resistors having values of 4 Ω, 6 Ω , and 8Ω are connected in series. Their equivalent resistance is ______.Group of answer choices18 Ω8 Ω6 Ω1.80 Ω
Answer:
18Ω
Explanation:
If the resistors are connected in series, the equivalent resistance is the sum of each resistance, so
Equivalent resistance = 4Ω + 6Ω + 8Ω
Equivalent resisteance = 18Ω
Therefore, the answer is 18Ω
Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one student was chosen at random, find the probability that the student got a B.
Answer:
20.45%
Explanation:
The probability that the student got a B is
[tex]\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%[/tex]Now, how many students are there in total?
The answer is
[tex]10+3+16+4+6+5=44\; \text{students}[/tex]How many students got a B?
The answer is
[tex]3+6=9\; \text{students}[/tex]therefore, the probability that the student has got a B is
[tex]\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%[/tex]Hence, the probability that a student has got a B is 20.45%
25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
Work = change in energy
Fd = 1/2 mv^2
F = 1/2 x 120 x 5^2 / 10
F = 150 N
If a student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m, then the average braking force would be 150 Newtons.
What is power?The rate of doing work is known as power. The Si unit of power is the watt.
Power =work / time
Work done by the braking force = change in kinetic energy
F × s = 1/2 × m × v²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force would be 150 Newtons.
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Physics
Hi dears how could we solve this question basically how to plug it in calculator
The new length of the steel bar is 11.65 cm.
What is linear expansion?
Linear expansion can be defined as the increase in length of a material due to increase in the temperature of the material.
Mathematically, the linear expansion of a material is given as;
ΔL = L₀αΔT
where;
ΔL is the change in length or increase in lengthL₀ is the initial length of the steelα is the coefficient of linear expansion of steel = 11 x 10⁻⁶/⁰CΔT is change in temperatureThe change in the length of the steel is calculated as follows;
ΔL = (11.5 cm) x (11 x 10⁻⁶/⁰C) x (1221 ⁰C - 22 ⁰C)
ΔL = 0.152 cm
The new length of the steel bar is calculated as follows;
L = ΔL + L₀
L = (0.152 cm) + (11.5 cm)
L = 11.65 cm
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Examine the heating curve for water below. Answer each question andcomplete the table to review your understanding of heating curves
Explanation:
When the temperature is increasing, the kinetic energy change, and when the graph is horizontal (the water is changing phases), the potential energy changes.
So at A, C, and E, there is a change in kinetic energy and at B and D there is a change in potential energy.
Then, we use a specific heat capacity of 4.2 J/g°C when the water is at A, C, and E. We use latent heat of 334 J/g for melting and latent heat of 2260 J/g for evaporation.
Answer:
Therefore, the complete table is
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision.. A)each car has half the impulse of before B)each car has double the impulse of before C)each car have the same impulse D)each car has an impulse in ratio to its mass.
We are given the following information.
The collision is elastic.
Car B has twice the mass of car A.
Recall that in an elastic collision, the momentum and the kinetic energy are conserved.
Impulse is basically the change in momentum.
Since car B has 2 times the mass of car A, the momentum of car B will be 4 times the momentum of car A.
This means that the impulse of car B will be greater than the impulse of car A.
Option D says that each car has an impulse in ratio to its mass meaning that a car with a larger mass will have a larger impulse and vice versa.
Therefore, we can conclude that after the collision, each car has an impulse in ratio to its mass.
What is the momentum of a 934 kg car moving 10 m/s?
ANSWER:
9340 kg*m/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 934 kg
Speed (v) = 10 m/s
The formula to calculate the momentum is as follows:
[tex]\begin{gathered} p=m\cdot v \\ \text{ we replacing} \\ p=934\cdot10 \\ p=9340\text{ kg*m/s} \end{gathered}[/tex]The momentum of the car is 9340 kg*m/s.
