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3 years ago

Which statement best compares potential and kinetic energy?

Physics

1 answer:

horrorfan [7]3 years ago

3 0

Answer:

Explanation:

Kinetic energy is basically energy while in motion, while potential is energy stored in a still object. When that still object starts moving the stored energy slowlyt decreases while turning into kinetic energy allowing the kinetic energy to increase.

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A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

vovikov84 [41]

In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

KE1 = KE2

The kinetic energy of the system before the collision is solved below.

KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s.

6 0

3 years ago

Which of the following is an effect of the electric force?

stich3 [128]

As we know that two charges exert force on each other when they are placed near to each other

The force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1, q_2$ = two different point charges

r = distance between two point charges

also we know that two similar charges always repel each other while two opposite charges always attract each other

so here correct answer would be

<em>A. A positive and negative charge attract each other.</em>

6 0

3 years ago

Read 2 more answers

When grocery shopping, the mass of the cart changes as you start to fill up your cart. How does the change in mass of your cart

Tasya [4]

Answer:

Explanation:

4 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

4 years ago

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob

lorasvet [3.4K]

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N$

Therefore, the force of attraction between the two charges is 3.55 N.

3 0

3 years ago