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A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

4 years ago

The heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water

Naily [24]

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

5 0

3 years ago

¿que es la energia interna?

Vladimir [108]

Answer:

La energía interna es el resultado de la contribución de la energía cinética de las moléculas o átomos que lo constituyen, de sus energías de rotación, traslación y vibración, además de la energía potencial intermolecular debida a las fuerzas de tipo gravitatorio, electromagnético y nuclear.

Explanation:

4 0

4 years ago

Read 2 more answers

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0

3 years ago

The weight of an astronaut plus his space suit on the moon is only 291 n. how much (in n) do they weigh on earth? (the accelerat

Oksanka [162]

Weight on the Moon = 291 N.
W = g · m, where m stays for the mass and on the Moon g = 1.67 m/s²
291 N = 1.67 m/s² · m
m = 291 kg m / s² : 1.67 m/s²
m = 174.25 kg
Weight on Earth = 9.81 m/s² · 174.25 kg = 1,709.4 N
Answer:
The weight of an astronaut on Earth is 1,709.4 N.

6 0

3 years ago