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2 years ago

6

a force of 25.0 newtons is applied so as to move a 5.0 kg mass a distance of 20.0 meters. How much work was done? with work plea

se

1 answer:

miv72 [106K]2 years ago

4 0

First we need to know the equation of work
work=force times distance
w= 25 N times 20 meters
w= 500 joules

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Any ferrous metal object within or near the mri magnet has the potential of becoming a projectile. this is commonly referred to

PtichkaEL [24]

This phenomena is referred to as metal projectile injury/effect. MRI stands for magnetic resonance imaging. As expected, your body is subjected to a magnetic field. This is a technique to look inside your body without cutting it open. Metals should not be placed anywhere near an MRI because it will cause the metals to follow the direction of the magnetic field.

4 0

2 years ago

A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial

puteri [66]

Answer:

A) vertically upward

Explanation:

Since the tyre is rotating with uniform angular speed and moving with constant linear speed

So as soon as a small stone is stuck into the groove of the tyre the speed of the stone is same as that of the tyre

so now we can say that stone will start revolving with the tyre of the car at constant angular speed and moving with uniform speed also

so here just after that the tangential acceleration of the stone must be zero while radial acceleration must be towards the center of the tyre given as

$a_c = \omega^2 R$

so we will have direction of net acceleration is towards its center so correct answer will be

A) vertically upward

7 0

2 years ago

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

olganol [36]

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, $a=\frac{F}{m}$

Final velocity , $v=a+ut$

Substitute the values

$v=0+\frac{F}{m}t=\frac{F}{m}t$

We know that

Momentum, p=mv

Using the formula

$p=m\times \frac{F}{m}t=Ft$

6 0

2 years ago

Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir

Svet_ta [14]

Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)

3 0

2 years ago

A single slit diffraction experiment performed with an argon laser of wavelength 454.6 nm produces a pattern on a screen with da

tankabanditka [31]

To develop this problem it is necessary to apply the concepts related to Interference and the Two slit experiment.

Young's equation defines the separation between fringes this phenomenon as,

$d = \frac{\lambda D}{a}$

Where,

$\lambda =$ Wavelength

d = Separation between fringes

a = Slit width

D = Distance between the slits and screen

Then for the two case we have

$d_1 = \frac{\lambda D}{a_1}$

$d_2 = \frac{\lambda D}{a_2}$

Calculating the new separation between the fringes would be

$\frac{d_2}{d_1} = \frac{\frac{\lambda D}{a_1}}{\frac{\lambda D}{a_2}}$

$\frac{d_2}{d_1} = \frac{\lambda_1}{\lambda_2}$

$d_2= \frac{\lambda_1}{\lambda_2} d_1$

We have then,

$d_2 = \frac{632.9*10^{-8}}{454.6*10^{-9}} 10*10^{-3}$

$d_2 = 13.9mm$

Therefore the correct answer is c.

7 0

2 years ago