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likoan [24]
3 years ago
13

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its _____.

Physics
1 answer:
olga2289 [7]3 years ago
4 0

Answer:

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>

Explanation:

By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.

This is given by :

F=m\ a

Where F represents force applied on the object , m represents mass of the object and a represents the acceleration.

In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.

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lbvjy [14]

Answer:

D for the first one and A for the second.

Explanation:

Hurricanes are usually caused by warmer waters.

Wind is created due to the movement of air because of the temperatures.

3 0
3 years ago
The following are the Earth–Sun distance at the equinoxes and solstices: March equinox 149.0 million km June solstice 152.0 mill
Mice21 [21]

Answer:

During <u>winter (late December/early January)</u> the Earth is closest to the Sun and during <u>summer (late June/early July)</u> the Earth is farthest from the Sun.

Explanation:

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On the other hand, the earth is farthest from the sun during the time of summer season, mostly in late June or early July.

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5 0
3 years ago
If a man has a made of 83 kilograms on earth, what will the force of gravity on his body on the moon
Salsk061 [2.6K]

the man would only weigh 13.7 kg on the moon

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7 0
3 years ago
Read 2 more answers
If temperature and the number of particles remain constant, and increase in pressure will cause volume to
omeli [17]
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5 0
3 years ago
A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V
Stels [109]

Answer:

2.6 kilo Ohm

Explanation:

Capacitance, C = 591 μF = 591 x 10^-6 F

Vo = 88.5 V

V = 11.9 V

t = 3.09 s

Let the resistance is R.

V = V_{0}e^{\frac{-t}{RC}}

\frac{11.9}{88.5} = e^{\frac{-t}{RC}}

0.135 = e^{\frac{-t}{RC}}

Take natural log on oth the sides

ln 0.135 = - 3.09 / RC

RC = 1.545

R = 1.545 / ( 591 x 10^-6)

R = 2614.2 ohm

R = 2.6 kilo Ohm

Thus the resistance is  2.6 kilo Ohm.

8 0
3 years ago
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