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trapecia [35]
3 years ago
8

An object with a mass of 2.3 kg has a force of 6.2 newtons applied to it. What is the resulting acceleration of the object?

Physics
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:2.7m/s^2

Explanation:

mass=2.3kg

Force=6.2Newton

Acceleration=force ➗ mass

Acceleration=6.2 ➗ 2.3

Acceleration=2.7m/s^2

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4 0
3 years ago
Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and
Zigmanuir [339]

Answer:

F=5.8\times 10^{8}\ N

F=35.57\times 10^{21}\ N

Explanation:

Given that

Intensity I

I= 1.39\ KW/m^2

Speed\ of \ light = 2.99\times 10^8\ m/s

Radius of earth,R = 6370 Km

We know that surface area of earth, A

A=\pi R^2

A=\pi (6370\times 10^3)^2\ m^2

A=1.27\times 10^{14}\ m^2

As we know that pressure due to intensity given as

P=\dfrac{I}{V}

V =Velocity of light

V=3\times 10^8\ m/s

P=\dfrac{1.39\times 1000}{=3\times 10^8}

P=4.6\times 10^{-6}\ Pa

We know that force F

F = P .A

F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N

F=5.8\times 10^{8}\ N

b)Gravitational force F

F=G\dfrac{m.M}{r^2}

M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg

r =1.5\times 10^{11}\ m

G =6.67\times 10^{-11}Nm^2/kg^2

So F

F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}

F=35.57\times 10^{21}\ N

7 0
3 years ago
Read 2 more answers
A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same
emmainna [20.7K]
This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal. 
                                   momentum = mass x velocity
            (1,600 kg)(16 m/s) + (1.0x10^3 kg)(10 m/s) = (1600 + 1000 kg)(x)
The value of x is 13.69 m/s. Thus, their final speed is approximately letter D. 14 m/s. 
6 0
4 years ago
A race car starting from rest accelerates uniformly at a rare of 4.90 meters per second^2. What is the cars speed after it has t
Vesnalui [34]

From the law of Galileo Galilei  :v²=v₀²+2ad we take the speed

v²=0+2*4.90*200=1960=>v=√1960=44.27 m/s




4 0
3 years ago
How do I solve for this?
vova2212 [387]
I can’t see anything
7 0
2 years ago
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