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3 years ago

ASAP PLS HELP What is chemical potential energy?

Physics

1 answer:

mash [69]3 years ago

5 0

Answer:

chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, e.g. in a chemical reaction or phase transition.

Explanation:

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An object moving north with an initial velocity of 14m/s accelerates 5m/s squared for 20 seconds. What is the final velocity of

Gemiola [76]

Since it moves 5 m/s faster every second, after 20 seconds it's moving 100 m/s faster than when it started speeding up.

If it was moving at 14 m/s when the acceleration began, it's moving at 114 m/s at the end of the 20 seconds. Its velocity is <em>114 m/s North.</em>

That's 255 mph !

4 0

3 years ago

Plz help i need help on my homework

tatiyna

The resultant force is 4 N to the right.

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4 years ago

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Matt is driving down 7th street. He drives 150 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe

Olin [163]

(150 meters) / (18 seconds) = 8.33 meters per second

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4 years ago

A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the t

cricket20 [7]

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

$F = \mu mg$

so we have

$F = ma = \mu mg$

now the acceleration is given as

$a = \mu g$

$a = (0.32)(9.8) = 3.14 m/s^2$

5 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago