What energy is present when a magnet is moving?

Umnica [9.8K]

Since the baseball is half the weight of the softball, the baseball has to go twice as fast as the softball.

So the best answer would be D

5 0

3 years ago

Read 2 more answers

A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

$\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s$

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

$\epsilon = \frac{d \phi}{dt}$

$\int E.dl = \frac{d(BA)}{dt}$

$E(2\pi r)= \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} \frac{dB}{dt}$

electric field at point P_1 as follow

$E = \frac{r}{2} \frac{dB}{dt}$

$E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}$

$E = 6.3\times 10^{-6} V/m$

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}$

a = 603.59 m/s^2

5 0

2 years ago

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

6 0

2 years ago

I will be so thankful if u answer correctly!!

olga_2 [115]

<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force $F_{r}$ is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F - $F_{r}$

m = 50kg

a = 0 [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F - $F_{r}$ = m x a

F - $F_{r}$ = 50 x 0

F - $F_{r}$ = 0

F = $F_{r}$ -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force $F_{r}$ is opposing the movement of the box.

∑F = 1.5F - $F_{r}$

m = 50kg

a = 0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F - $F_{r}$ = m x a

1.5F - $F_{r}$ = 50 x 0.1

1.5F - $F_{r}$ = 5 ---------------------(iii)

<em>Substitute </em> $F_{r}$ <em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5

0.5F = 5

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

4 0

2 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

2 years ago