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Gwar [14]
3 years ago
14

Which of the following best demonstrates the effect of static friction? A. A person dropping a ball to the ground. B. A person p

ushing a couch and it slowly sliding across a floor C. A person pushing a couch without being able to move it D. A person moving a ball through a stream of water
Physics
1 answer:
Dominik [7]3 years ago
6 0

Answer:

C.

Explanation:

A person pushing a couch will face resistive force of friction . When resistive force is greater then his force of effort , couch will not move. This force is static friction because the couch is stationary. When the force of effort is increased , magnitude of static friction also increases keeping the couch stationary. The ability of the static friction to increase its magnitude is limited by a maximum value beyond which the couch starts moving. The static friction is then converted into kinetic friction.

In rest of the three cases object is already moving so kinetic friction is in action.

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Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th
tatiyna

Answer:

b) -10 m/s

Explanation:

In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.

3 0
3 years ago
Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,
andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

<u>D = - 0.5i - 0.25j</u>

Therefore,

<u>Dx = -0.5</u>

<u>Dy = -0.25</u>

8 0
3 years ago
Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave
mylen [45]

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

7 0
3 years ago
A 20-kilogram child is riding on a 10-kg sled over a frictionless icy surface at 8.0 meters per second. Calculate the kinetic en
Veseljchak [2.6K]

Answer:

K = 960 J

Explanation:

Given that,

Mass of a child = 20 kg

Mass of a sled = 10 kg

Speed of child on sled = 8 m/s

We need to find the kinetic energy of the sled with the child.

The total mass of child and the sled = 20 kg + 10 kg

= 30 kg

The formula for the kinetic energy of an object is given by :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 30\times (8)^2\\\\K=960\ J

Hence, the kinetic energy of the sled with the child is 960 J.

6 0
3 years ago
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 pis
Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0
3 years ago
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