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3 years ago

How long would it take a machine to do 5.000

Physics

1 answer:

Gala2k [10]3 years ago

6 0

Answer:

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

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A 50 kg woman, riding on a 10 kg cart, is moving east at 5.0 m/s. the woman jumps off the cart and hits the ground running at 7.

pav-90 [236]

To answer this question, we will use the law of conservation of momentum which states that:
(m1+m2)Vi = m1V1 + m2V2 where:
m1 is the mass of the woman = 50 kg
m2 is the mass of the cart = 10 kg
Vi is the initial velocity (of woman and cart combined) = 5 m/sec
V1 is the final velocity of the woman = 7 m/sec
V2 is the final velocity of the cart that we need to calculate

Substitute with the givens in the above equation to get the final velocity of the cart as follows:
(50+10)(5) = (50)(7) + (10)V2
10V2 = -50
V2 = -5 m/sec
Note that the negative sign indicates that the cart is moving in an opposite direction to the others.

4 0

3 years ago

A battery and a resistor are wired into a circuit. The resistor dissipates 0.30 W. Now two batteries, each identical to the orig

nata0808 [166]

Answer:

1.2 W

Explanation:

Let the value of resistance be R , emf of battery be E

current i = E / R

power = i² R

= E² x R / R²

= E² / R

Given

E² / R = .30

when two batteries are connected

Total emf = 2E

current = 2E / R

Power = ( 2E / R )² x R

= 4 E² / R

= 4 x .30

= 1.2 W .

7 0

3 years ago

A hollow sphere of radius 1.88 m is in a region where the electric field is radial and directed toward the center of the sphere.

Marta_Voda [28]

Answer:

1003.9 Nm²/C

Explanation:

Parameters given:

Radius of sphere, r = 1.88 m

Electric field at surface of sphere, E = 22.6 N/C

Electric Flux, Φ, is given as:

Φ = E * A

Where A = Surface area

Surface Area of sphere = 4 * pi * r²

= 4 * pi * 1.88²

= 44.4 m²

Φ = 22.6 * 44.4

Φ = 1003.9 Nm²/C

8 0

3 years ago

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:

$W=\Delta K=K_f -K_i$