Ask question

Ask question

All categories

2 years ago

13

When work is done by an applied force, the object's energy will change. In this Interactive, does the work cause a kinetic energ

y change or a potential energy change?

1 answer:

Alex73 [517]2 years ago

6 0

Answer:

Explanation:

According to the work energy theorem, the work done by all the forces by a body is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

W = Final kinetic energy - initial kinetic energy

So, work causes change in kinetic energy.

You might be interested in

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

2 years ago

(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c

umka2103 [35]

Answer:

$E=\frac{k\,Q}{d^2}$

Explanation:

The strength of an electric field E produced by a single charge Q at a distance d from it is given by the formula: $E=\frac{k\,Q}{d^2}$ , where K represents the Coulomb constant.

Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

$F_C=k\frac{Q_1\,Q_2}{d^2}$

but modified with only one charge showing in the numerator of the expression.

8 0

2 years ago

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli

Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: $sin \theta=\frac{m\lambda}{\alpha}$

Given $m=3$ , $\lambda=6.33\times 10^-^7$ and $\alpha=0.00015$ :

#Substitute our variable values in the minima equation to obtain $\theta$ :

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

#draw a triangle to find the relationship between $\theta, y \$ and $L$ .

$tan(\theta)=y/L$ #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b. To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$ , where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$ =4.5cm

Then from the equation subtract $y_3$ from $y$ :

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm

8 0

2 years ago

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre

alex41 [277]

Answer:

The simplified expression is $M = \frac{v^2 r}{G}$

Explanation:

From the question we are told that

$M = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }$

So simplifying we have

$M = \frac{m v^2}{r} * \frac{r^2 }{ mG }$

$M = \frac{v^2 r}{G}$

Thus the simplified formula is $M = \frac{v^2 r}{G}$

3 0

2 years ago

Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe

levacccp [35]

Answer:

Obviously Lengthen... $T = 2\pi \sqrt{L/g}$ or $g = 4\pi ^{2} L/g$

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

4 0

2 years ago