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3 years ago

A 2000 kg car is moving at 1.5 m/s. What is its kinetic energy?*

Physics

2 answers:

Rom4ik [11]3 years ago

6 0

Its E,
K.E = 1/2mv^2
K.E = 1/2(2000)(1.5)^2
K.E = 2250 J

Dmitrij [34]3 years ago

3 0

Answer is B 3000 J hope I could help

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A person absentmindedly walks off the edge of a tall cliff. They will fall 50 m into either

kifflom [539]

The man can survive, and he lands 12.1 m from the base of the cliff (into the lake)

Explanation:

The motion of the person is equivalent to the motion of a projectile, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion, to find the time it take for the person to reach the ground level. We can use the following suvat equation:

$s=ut+\frac{1}{2}gt^2$

where

s = 50 m is the vertical distance covered (the height of the cliff)

u = 0 is the initial vertical velocity

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

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(50)}{9.8}}=3.19 s$

The person moved horizontally at a constant speed of

$v_x = 3.8 m/s$

So, the horizontal distance covered by the man during his flight will be

$d_x = v_x t = (3.8)(3.19)=12.1 m$

So, the man will land on the lake (which starts from 12 m), so he can survive.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

3 years ago

What is the speed of sound dependent on?

mojhsa [17]

Depends on the elasticity and density of the medium through what it is traveling <span> </span>

5 0

3 years ago

A car starts from 0 m along a road and accelerates at 0.5 m/s^2 to the right. A second car starts from 1000 m along the road and

Brrunno [24]

Answer:

e) 31.6 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow s_1=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_1=\frac{1}{2}0.5t^2\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s_2=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_2=\frac{1}{2}1.5t^2\ m$

$s_1+s_2=1000$

$\\\Rightarrow 1000=\frac{1}{2}0.5t^2+\frac{1}{2}1.5t^2\\\Rightarrow 1000=\frac{0.5t^2+1.5t^2}{2}\\\Rightarrow 1000=\frac{2t^2}{2}\\\Rightarrow 1000=t^2\\\Rightarrow t=\sqrt{1000}\\\Rightarrow t=31.6\ s$

Time taken by the cars to meet 31.6 seconds.

5 0

3 years ago

E) Thermal energy is released during

vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

$Q_{net,in} - W_{net, out} = \Delta U$ (1)

Where:

$Q_{net, in}$ - Net input heat, measured in joules.

$W_{net, out}$ - Net output work, measured in joules.

$\Delta U$ - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then $\Delta U < 0$ , which is caused by three scenarios:

(i) $Q_{net,in} < 0$ , $W_{net, out} < 0$ , $|Q_{net,in}|>|W_{net,out}|$

(ii) $Q_{net, in} > 0$ , $W_{net,out} > 0$ , $|Q_{net,in}|$

(iii) $Q_{net,in}< 0$ , $W_{net, out}>0$

In the case $Q_{net,in} > 0$ , $W_{net, out}$ , the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

$Q_{net, in} = \Delta U$ , $Q_{net, in} < 0$ (2)

If $Q_{net, in} < 0$ , then $\Delta U < 0$ . Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0

3 years ago

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea

Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Where,

L = Vibrating length string

T = Tension in the string

$\mu =$ Linear mass density

At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

$f_1$ = First frequency

$f_2$ = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

$f \propto \sqrt{T}$

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

$n = f_2-f_1$

$f_2 = n+f_1$

Replacing $3/s$ for n and 202Hz for $f_1,$

$f_2 = 3/s + 202Hz$

$f_2 = 3/s(\frac{1Hz}{1/s})+202Hz$

$f_2 = 206Hz$

The frequency of the tightened is 205Hz

7 0

3 years ago