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Natali5045456 [20]
3 years ago
5

A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per seco

nd east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds?
(1) 2.0 m/s east
(2) 10.0 m/s east
(3) 2.0 m/s west
(4) 10.0 m/s west
Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

(2) 10.0 m/s east

Explanation:

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5)A 0.50 kg hockey puck is at rest on ice when you hit it with a hockey stick, applying a force of 100 N for
mojhsa [17]

Answer:

F t = m Δv         impulse delivered = change in momentum

Δv = 100 * .1 / .5 = 20 m/s     original speed of puck

KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J     initial KE of puck

E = μ m g d        energy lost by puck

Ff = μ m g = m a      deceleration of puck due to friction

a = μ  g = 9.8 * .2 = 1.96 m/s^2

v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s     speed of puck on striking box

m v2 = M V       conservation of momentum when puck strikes box

V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s     speed of box after collision

KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J     KE of box after collision

KE = μ M g d     energy lost by box in sliding distance d

d = 23.3 / (.3 * .8 * 9.8) = 9.91 m     distance box slides

7 0
2 years ago
I have a question regarding friction in rolling without slipping.
Solnce55 [7]

Explanation:

They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.

(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)

In this case, you can't ignore friction because the disk wouldn't roll without it.

As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.

(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)

Hopefully this helps clear up the misunderstanding for you.

4 0
3 years ago
Suppose that a car performs a uniform acceleration of 0.42 m/s from rest to 30.0 km/h in the first stage of its motion (From poi
kifflom [539]

Answer:

a)Total distance = 399. 5 m

b)Total time =51.51 sec

c)Average speed = 7.75 m/s

Explanation:

For A to B:

S=ut+\dfrac{1}{2}at^2

v^2=u^2+2as

v= u + at

8.33 = 0.42 x t

t=19.83 sec

1 Km/h=0.27 m/s

30 Km/h=8.33 m/s

8.33^2=2\times 0.42\times s

s=82.6 m

For B to C

V= 8.33 m/s

s= V x t

s=8.33 x 30

s=249.9 m

For C to D

S=ut-\dfrac{1}{2}at^2

v= u - at

Final speed v=0

So

s=v x t/2

7= 8.33 x t/2

t=1.68 sec

Total distance = 82.6 + 249 .9 +7

Total distance = 399. 5 m

Total time = 19.83 + 30 + 1.68

Total time =51.51 sec

Average speed =Total distance/Total time

Average speed = 399.5/51.5

Average speed = 7.75 m/s

6 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed v of the geosynchronous satellite is given by:

v=\frac{2\pi R}{T}

Because 2\pi R is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

F_c=\frac{mv^{2} }{R}

If we substitute the expression for v in this formula, we get:

F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}

Since the centripetal force is the gravitational force F_g between the satellite and the Earth, we know that:

F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }

Where G is the gravitational constant (G=6.67*10^{-11} Nm^{2}/kg^{2}) and M is the mass of the Earth (M=5.97*10^{24}kg). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0
4 years ago
Which statement describes an essential characteristic of data in an experiment?
MAVERICK [17]
The whole of the data that you have researched on
3 0
4 years ago
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