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noname [10]
3 years ago
12

Is a force pushing up on an object as gravity is pulling down?

Physics
1 answer:
aliya0001 [1]3 years ago
6 0

Answer:

Logically yes, because Newton's Third law state "When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."

If force wasn't pushing up then neither gravity is pulling down.

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A boat moves through the water with two forces acting on it. One is a 1,800-N forward push by the water on the propeller, and th
baherus [9]

(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.

(c)Velocity at the end of that time will be 8.5 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f  = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f  = 85 \ m

c)

The velocity at the end of that time is found as;

\rm  v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5 m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

#SPJ1

8 0
2 years ago
two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change
Verdich [7]
The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
5 0
3 years ago
15) Very cold climates occur at Earth's North and South Poles because the polar regions
Gwar [14]

Because the polar regions receive low-angle insolation.

Insolation is the amount of solar radiation received by a given area. The Sun is always low on the horizon. The low Sun angle makes the beam of solar radiation to travel a longer distance from upper troposphere to reach earth's surface as compared to when it is directly overhead. In this case, the radiations are scattered and reflected more by the atmosphere and spread over a larger area. Thus, the intensity of solar radiation is very less at polar regions than near the equatorial region. This is the reason of very cold climates at polar regions.

3 0
3 years ago
Read 2 more answers
The moon Phobos orbits Mars
podryga [215]

Explanation:

For a circular orbit v= \sqrt{\frac{G.m}{r} } with G = 6.6742 × 10^{-11}

Given m = 6.42 x 10^23 kg and  r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

3 0
3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that
Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

T = \frac{t}{n}

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

T = \frac{135\ s}{98}

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

T = 2\pi \sqrt{\frac{l}{g}}

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}

<u>g = 11.2 m/s²</u>

3 0
3 years ago
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