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2 years ago

When an object moves, where does the energy come from?

1 answer:

5 0

Ans: Kinetic and potential energies are found in all objects. If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick.

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In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial velocity of each object

$v_f$ = Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

$m_2v_2 = (m_1+m_2)v_f$

$(0.42)(21) = (90+0.42)v_f$

$v_f = 0.0975m/s$

Therefore the velocity right after catching the ball is 0.0975m/s

8 0

3 years ago

A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?

12345 [234]

The block is made of A) Tin, as its specific heat capacity is $0.225 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by $\Delta T$ , according to the following equation :

$Q=mC_s \Delta T$

where $C_s$ is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

$Q = 450 J$ is the amount of energy supplied to the block

$\Delta T = 1^{\circ}C$ is the change in temperature of the material

Solving the equation for $C_s$ , we can find the specific heat capacity of the unknown sample:

$C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)$

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

7 0

2 years ago

A ball is held at rest at the top of a hill. The ball is then released and starts rolling down the hill. At the bottom it reache

Alex17521 [72]

Answer:

Gravitational Potential Energy

Explanation:

a ball is held rest at the top of hill

gravitational potential energy will store due to its height

it. and body will start move downward and its potential energy will convert into kinetic energy due to motion of body

at the ground level it will stop and potential energy will became zero and kinetic energy get convert into internal energy due to collisions

3 0

2 years ago

WILL MARK BRAINLIEST!!!!

zloy xaker [14]

Answer:

Wave model

Explanation:

7 0

2 years ago

Read 2 more answers

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

2 years ago