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1 year ago

Base your answers to questions 4 and 5 on the informa- tion below

1 answer:

3 0

If the sphere is positively charged, the positive rod is repelled by the sphere while the negative rod is attracted by the sphere.

<h3>What is an electrical charge?</h3>

An electrical charge can be positive or negative. From the laws of electrostatics, unlike charges attract while like charges repel. As such the effect observed when the rods are individually brought near the sphere will decide the charge on the sphere.

If the sphere is neutral, there is no effect observed when each rod is brought near the sphere. If the sphere is positively charged, the positive rod is repelled by the sphere while the negative rod is attracted by the sphere.

Learn more about electrostatics: brainly.com/question/9774180

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A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis

Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

$M_{before} = M_{after}$

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

$(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})$

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 = velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

3 0

2 years ago

Describe how water may be treated before people use it

brilliants [131]

Water is treated by purifying it by adding slaked lime or potash alum

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2 years ago

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Does a stone lying on the ground have stored energy

Vesnalui [34]

Yes, It has a stored energy in that stone.

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2 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

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3 years ago

Which scientist did this?

Colt1911 [192]

You answer would be C. Ernest Rutherford.

4 0

2 years ago

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