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2 years ago

Question 7 of 10

1 answer:

4 0

Kinetic friction (also referred to as dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion.
https://www.khanacademy.org › stat...
Static and kinetic friction example (video) | Khan Academy

Answer a would be static friction
Answer b is fluid friction
(Air resistance is fluid friction. Fluid friction is the friction experienced by objects which are moving in a fluid and the air is a fluid.)
Answer c is static friction
ANSWER D IS KINETIC FRICTION

Hope this helps :D

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The weight (W) of an object is the product of its mass (m) in kilograms and g, the acceleration due to gravity in meters/second2

Nookie1986 [14]

The weights in newtowns for the given masses are

<span> masses 22.1, 33.5, 41.3, 59.2, 78
weights 216.58N 328.3N 404.74N 580.16N 764.4N

e.g, for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>

7 0

3 years ago

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

The entropy change of the rod will be zero.

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

3 years ago

Which of the following factors affects the pressure of an enclosed gas?

melamori03 [73]

It would be d all of the above

5 0

3 years ago

Read 2 more answers

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

3 years ago

Is how high or low sound is

JulsSmile [24]

Answer:

How high the sound or how low the sound is depending on the pitch of the sound which in this case the frequencies of the sound. The higher of the frequencies , the higher of the pitch but it has the shortest length of wave (λ). That's why AM radio have a longer range but bad audio quality than FM radio that have better audio quality with shorter range.

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3 years ago