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4 years ago

Work & Power Problems

Physics

1 answer:

Artyom0805 [142]4 years ago

6 0

Answer:

1. Unit for work is measured in Joules.

Formula for work is power * time taken

2 The unit for force is Newton

A work done in lifting a jar of water

3. The unit for measuring power is Joules.

The formula for measuring power is work done/time taken

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A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

The wave shown below is produced in a rope.

Kay [80]

Answer:

<h3>B) A dot vector drawn up and vector drawn down.</h3>

Explanation:

Brainliest?

6 0

3 years ago

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After the collapse of a nebular cloud, atoms begin gravitating together to form a condensed center. What happens next in the sta

ELEN [110]

Well, first of all, I don't think "After the collapse of a nebular cloud ..."
is the first time that "atoms begin gravitating together". Seems to me like
that's what was going on all the time, and it's what caused the nebular cloud
to collapse in the first place.

In any case, once the pressure and temperature at the center get high enough,
you get "ignition" of nuclear fusion, and that's when you first have a "star".

8 0

4 years ago

Read 2 more answers