All categories

2 years ago

What happens to your kinectic energy when your gravitational potential energy decreases?

Physics

1 answer:

leonid [27]2 years ago

3 0

Answer:

As r decreases, we lose gravitational potential energy - in other words, U G U_G UGU, start subscript, G, end subscript becomes more negative. Because energy is conserved, the velocity must increase, resulting in an increase in kinetic energy.

Explanation:

You might be interested in

Hurricanes are considered____ because they lose power over cool waters or land A. Short-lived B. Heat engines C. Weak

klasskru [66]

My best guess would be heat engines

7 0

3 years ago

Read 2 more answers

I'm sorry if any one seen that post about me saying I'm done with life and I don't want to be here it was my little cuzin @vicky

solniwko [45]

I did not see the post. But if she actually needs help She can dm me. keep y'all's head up❤

8 0

2 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago

A person is initially driving a car east down a straight road. the magnitude of the instantaneous acceleration is decreasing wit

Alexxandr [17]

By definition, acceleration is the change in velocity per change of time. As time passes by, the time increases in value. So, when the acceleration is decreasing while the time is increasing, then that means that the change of velocity is also decreasing with time. So, optimally, the initial velocity and the velocity at any time are very relatively close to each other,

4 0

2 years ago

Read 2 more answers

What is the mechanical advantage of a nail puller where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the oth

Mandarinka [93]

Answer:

Explanation:

Mechanical Advantage is the ratio of the distance of the input load (Li)from the pivot to the output load applied to the pivot(Lo)

MA = Li/Le

Given;

Li = 45cm

Lo = 1.8cm

MA = 45/1.8

MA = 25

Hence the mechanical advantage is 25

Also MA is expressed in terms of the force ratio which is the ratio of the Load to the effort applied.

MA = Load/Effort

Given

Load = 1250N

MA = 25

Effort = ?

Substitute

25 = 1250/Effort

Effort = 1250/25

Effort = 50N

Hence the minimum force exerted on the load is 50N

3 0

2 years ago