Calculate the gravitational potential energies of the melon and the pomegranate. Which one has potential energy

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

hodyreva [135]

Answer : The correct option is, (c) $3.7\times 10^2J/^oC$

Explanation :

First we have to calculate the energy or heat.

Formula used :

$E=V\times I\times t$

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

$E=(3.6V)\times (2.6A)\times (350s)$

$E=3276J$

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

$C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}$

where,

C = heat capacity of the calorimeter

$T_{initial}$ = initial temperature = $20.3^oC$

$T_{final}$ = final temperature = $29.1^oC$

Now put all the given values in this formula, we get:

$C=\frac{3276J}{(29.1-20.3)^oC}$

$C=372.27J/^oC=3.7\times 10^2J/^oC$

Therefore, the heat capacity of the calorimeter is, $3.7\times 10^2J/^oC$

7 0

3 years ago

At what distance will the weight of a body be halved , Earth's radius=6.4×10^6

Sholpan [36]

A body of mass m has weight

F = GMm/r²

on the surface of the Earth, where G is the universal gravitational constant, M is the mass of the Earth, and r is it's radius.

If the weight is to be halved, then we have

1/2 F = 1/2 GMm/r² = (1/√2)² GMm/r² = GMm/(√2 r²)

so the distance between the body and the planet's center needs to be

√2 × 6.4 × 10⁶ m ≈ 9.1 × 10⁶ m

5 0

3 years ago

If the total momentum of a system is changing:

DENIUS [597]

Answer:

(d) a net external force must be acting on the system

Explanation:

Momentum is given as the product of mass and velocity.

P = MV

According to Newton's second law of motion, " Force applied to a body (system) is directly proportional to the rate of change of momentum of the body (system) which takes place in the direction of the applied force (external force).

F ∝ΔMV

Therefore, If the total momentum of a system is changing, a net external force must be acting on the system.

(d) a net external force must be acting on the system

3 0

3 years ago

If velocity is positive, which woul

shusha [124]

Answer:

C. An inital volocity that is faster than the final volocity

Explanation:

.

5 0

3 years ago