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3 years ago

How much work is done if a force of 50 N is used to move a footstool 3 M?

Physics

1 answer:

Nuetrik [128]3 years ago

5 0

Answer: 150J

Explanation:

Work done = force * distance

W = 50N * 3M

W= 150J

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"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

notka56 [123]

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

4 0

3 years ago

a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti

PolarNik [594]

Answer:

B=0.2T

Explanation:

given required solution

l=4m B=? F=BIL

i=0.5A B=F/IL

F=0.4N B=0.4N/0.5A*4m

B=0.4/2=0.2T

5 0

3 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

3 years ago

A ball starts from rest and undergoes uniform acceleration of 2.50m/s^2. What is the velocity of the ball 4s later?

erica [24]

Explanation:

Given:

v₀ = 0 m/s

a = 2.50 m/s²

t = 4 s

Find: v

v = at + v₀

v = (2.50 m/s²) (4 s) + 0 m/s

v = 10 m/s

4 0

3 years ago

Read 2 more answers

A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

nikdorinn [45]

Answer:

$\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

$I = 0.106 \ A$

Explanation:

From the question we are told that

The current is $I = 0.106 \ A$

The length of one side of the square $a = 4.60 \ cm = 0.046 \ m$

The separation between the plate is $d = 4.0 mm = 0.004 \ m$

Generally electric flux is mathematically represented as

$\phi_E = \frac{Q}{\epsilon_o}$

differentiating both sides with respect to t is

$\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=> $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

Here $\epsilon_o$ is the permitivity of free space with value

$\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=> $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=> $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

Generally the displacement current between the plates in A

$I = 8.85*10^{-12} * 1.1977 *10^{10}$

=> $I = 0.106 \ A$

3 0

2 years ago