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3 years ago

How much work is done by a student holding a book with a force of 5N and walking forward 10 m?

Physics

1 answer:

inysia [295]3 years ago

7 0

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 5 × 10 = 50

We have the final answer as

Hope this helps you

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A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

3 years ago

Suppose you have a 75-watt light bulb that you leave turned on for two minutes. How much energy

victus00 [196]

Answer:

9000 J

Explanation:

Convert minutes to seconds.

2 min = 120 s

Power = energy / time

75 W = E / 120 s

E = 9000 J

5 0

2 years ago

This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa

Montano1993 [528]

Answer:

$2.77287\times 10^{15}\ m$

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

$\nu$ = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

$P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s$

Photon intensity is given by

$i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m$

The distance is $2.77287\times 10^{15}\ m$

3 0

3 years ago

If the wagon travels 18.75 m, what is the work done on the wagon

Schach [20]

If the wagon travels 18.75 m, then the work done on the wagon is

(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .

The unit is Joules .

4 0

4 years ago

Help me with this please

creativ13 [48]

Answer:

i dont understand, can you put it in word form please.

Explanation:

4 0

3 years ago