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diamong [38]
3 years ago
8

What is the amount of charge when 13.5a is flowing for 2 1/2 hours

Physics
2 answers:
abruzzese [7]3 years ago
4 0

Answer:

Quick Answer:

a.  8.9Ω

b.  1.2×104 C

Explanation:

Illusion [34]3 years ago
4 0

Answer:

Answer provided in explanation

Explanation:

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I NEED PHYSICS HELP? THE QUESTION IS IN THE PIC
Marina CMI [18]

Speed of the block at the bottom of the incline: 5.42 m/s

The first part of the problem can be solved by using the law of conservation of energy. Since the ramp is frictionless, the initial gravitational potential energy of the block at the top of the ramp is converted into kinetic energy at the bottom:

mgh = \frac{1}{2}mv^2 (1)

where

m is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

h is the initial height of the block

v is the speed of the block at the bottom

The initial height of the block is equal to the height of the ramp, so

h=L sin \theta (2)

where

L = 3.00 m is the length of the ramp

\theta=30^{\circ} is the angle of the ramp

Substituting (2) into (1) and re-arranging the equation, we  find the speed

2gL sin \theta = v^2

v=\sqrt{2gL sin \theta}=\sqrt{2(9.8)(3.00)sin 30^{\circ}}=5.42 m/s

Coefficient of kinetic friction between the floor and the block: 0.3

In the second part of the motion, the block is slowed down by friction along the flat surface. According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the block:

W=\Delta K=K_f -K_i

where

W is the work done by friction

Kf is the final kinetic energy of the block, which is zero since the block comes to rest

K_i = \frac{1}{2}mv^2 is the initial kinetic energy of the block, where

m = 10.0 kg is the mass of the block

v = 5.42 m/s is its initial speed

Substituting into the equation, we find

W=-\frac{1}{2}mv^2=-\frac{1}{2}(10.0)(5.42)^2=-146.9 J

and the work is negative, since the direction of the force of friction is opposite to the direction of motion of the block.

Now we can rewrite the work as the product between the force of friction and the displacement of the block:

W=-F_f d = - \mu mg d

where

\mu is the coefficient of friction

d = 5.00 m is the displacement of the block

Solving for \mu,

\mu = - \frac{W}{mgd}=-\frac{-146.9}{(10.0)(9.8)(5.00)}=0.3

4 0
3 years ago
A 62.5 kg carpenter at a construction site plans to swing in a circular arc from one roof top to an adjacent roof at the end of
denis23 [38]

Answer:

    v = 9.45 m/s

Explanation:

given,

mass of the carpenter = 62.5 Kg

length of rope = 13.4 m

Capable of exerting force = 1034 N

centripetal force acting on the body

                   F = \dfrac{mv^2}{r}

                   F = \dfrac{62.5\times v^2}{13.4}

                  F =4.664 v²  N

Gravitational force on her =

                  F = m g

                  F = 62.5 x 9.81

                  F = 613.125 N

now,

4.664 v² + 613.125 = 1034

4.664 v² = 420.875

   v²  = 90.24

    v = 9.45 m/s

Maximum speed which she can tolerate = v = 9.45 m/s

7 0
3 years ago
1.Seema is the wife of and Rahul is the brother of Robin. Robin is the only uncle of Ram .What is Ram relation with Seema ?​
Phantasy [73]

Answer:

Ram is the son of Seema

3 0
3 years ago
Read 2 more answers
Why are water ripples, electromagnetic waves, and radio waves transverse?
Alexandra [31]

Answer:

If a transverse wave is moving in the positive x-direction, its oscillations are in up and down directions that lie in the y–z plane. Light is an example of a transverse wave. With regard to transverse waves in matter, the displacement of the medium is perpendicular to the direction of propagation of the wave.

Explanation:

4 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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