All categories

3 years ago

Technically, how would you know if any work was done on an object?

Physics

1 answer:

kherson [118]3 years ago

8 0

Answer:

through reading the question carefully and using yourself as an exampke

You might be interested in

At a constant temperature, the volume of a gas doubles when the pressure is reduced to half of its original value. This is a sta

djyliett [7]

Answer: Boyle’s law

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

As pressure is decreased to half, the volume is increased to doubled.

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$ (At constant pressure and number of moles)

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$ (At constant volume and number of moles)

Combined gas Law: combining the three laws:

$PV\propto T$

6 0

4 years ago

Read 2 more answers

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago

A fleeing student covered a distance of 210 meters in 35 seconds. what is the student's speed in meters/second? what is the stud

snow_lady [41]

Explanation:

Given that,

A student covered a distance of 210 meters in 35 seconds.

We need to find the student's speed in meters/second and also in meters/minute.

Speed, v = distance (d)/time (t)

So,

$v=\dfrac{210\ m}{35\ s}\\\\=6\ m/s$

We know that, 1 minute = 60 seconds

$6\dfrac{m}{s}=6\times \dfrac{m}{(\dfrac{1}{60})\ \text{minutes}}\\\\=360\ \text{meters/minutes}$

Hence, the student's speed is 6 m/s or 360 meters/minute.

6 0

3 years ago

The property of matter that resists changes in motion is called

Nadusha1986 [10]

intertia hope that helps

3 0

4 years ago

In the one pully system when you move the mass from the 20 cm mark to the 15 cm mark, it moves 5 cm. How far did you pull the st

drek231 [11]

Answer:

5 cm

Explanation:

4 0

3 years ago