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dimaraw [331]
3 years ago
11

Does the distance between two charged objects affect the force of attraction?

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

Both forces act along the line joining the objects like masses or charges. And both forces are inversely proportional to the square of the distance between the objects, this is known as the inverse-square law.

Explanation:

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An object with a mass of 25 kg object is moving at 30 m/s. What is the
Pani-rosa [81]

Answer: 750 kgm/s

Explanation:

Mass of object = 25 kg

Speed by which object moves =30 m/s. Linear momentum of the object = ?

Since momentum refers to the quantity of motion of the moving object,

Linear momentum = Mass x Speed

= 25kg x 30m/s

= 750 kgm/s

Thus, the linear momentum of the object is 750 kgm/s

7 0
3 years ago
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE PUT TRUE ANSWERS
SIZIF [17.4K]

Answer:

D

Explanation:

A is wrong bc best DNA will survive

B is wrong because organism adapt not only for food source but to be able to live in the environment as well

C is wrong because single celled organisms don't adapt

6 0
3 years ago
A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

db=1.606T

So initial magnetic field = 1.606-0.3= 1.306 T

5 0
3 years ago
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe
dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is  3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F_{hand

using equation for impulse in momentum

F_{hand × t = m( v - v₀ )

we substitute

F_{hand × 0.00265  = 1.75( 0 - 5.25 )

F_{hand × 0.00265  = 1.75( - 5.25 )

F_{hand × 0.00265  = -9.1875

F_{hand = -9.1875 / 0.00265  

F_{hand = -3466.98 N

Next we determine force on the leg F_{leg

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F_{leg = - F_{hand

we substitute

F_{leg = - ( -3466.98 N )

F_{leg = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N

5 0
3 years ago
True or false radiant energy spreads out from its source in all directions
Sati [7]
Well the Answer is True.
6 0
3 years ago
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