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inessss [21]
2 years ago
7

Calculate the force of gravity acting on an object that has a mass of 1.3 kilograms. The object is on Earth soyuz 9.8 m / S 2 fo

r the acceleration due to gravity
A 12.7 N
B 1270 N
C 0.13 N
D 7.5 N
Physics
1 answer:
aleksklad [387]2 years ago
4 0
A) 12.7 cause
weight = g. mass
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Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert
damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

F = k \frac{q\cdot q}{r^2}

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

q+\frac{q}{2}=\frac{3}{2}q

while the other charge will be

q-\frac{q}{2}=\frac{q}{2}

So, the new force will be

F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F

So, the force will decrease to 3/4 of its original value.

6 0
3 years ago
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
6 0
3 years ago
Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30
barxatty [35]
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
5 0
2 years ago
H<br>Ggghffyjfdudjhfhghggffghjjdxv
amid [387]

Answer:

H

Ggghffyjfdudjhfhghggffghjjdxv

Explanation: The law of copying

6 0
3 years ago
Monochromatic light of a given wavelength is incident on a metal surface. however, no photoelectrons are emitted. if electrons a
kifflom [539]
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
5 0
2 years ago
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