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3 years ago

A 2 kg box of taffy candy has 40j of potential energy relative to the ground. Its height above the ground is

Physics

2 answers:

lord [1]3 years ago

8 0

Answer:

2.04m

Explanation:

PE=mgh

h= PE/mg

h= 40/2(9.82)

h=40/19.64

h=2.04

Alona [7]3 years ago

7 0

Answer:

$20m$

Explanation:

Formula to find the potential energy is,

Potential Energy = mgh

Here,

m = mass

g = gravitational field

h = height

According to the question, we have to find the height.

Usually, we get g is $10ms^{-2}$

Let us solve for the answer now.

$P.E. =mgh\\40J =2kg*10ms^{-2} *h\\40 = 20h\\\frac{40}{20}= \frac{20h}{20} \\20m=h$

Hope this helps you.

Let me know if you have any other questions :-)

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

3 years ago

What should be angle between force and displacement to get the maximum work

aleksandrvk [35]

Answer: The angle between force and displacement should be θ = 90° for minimum work. The angle between force and displacement should be θ = 0° for maximum work.

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3 years ago

Question 2: Start-Up

yuradex [85]

Answer:

The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.

Explanation:

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3 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$