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Marina CMI [18]
3 years ago
11

If a ball is given an acceleration of 3.0 meters/second^2 while being pushed with a force of 0.75 newtons, what is the mass of t

he ball?
Physics
2 answers:
Anit [1.1K]3 years ago
8 0
F=ma is one form of Newton's second law
m=F/a
m=(3/4)x1/3
m=9/4=2.25kg
yawa3891 [41]3 years ago
7 0

F = m.a

m = F/a

m = 0.75/3

m = 0.25 Kg OR 250 g

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At a place where an object is thrown vertically downward with a speed of while a different object is thrown vertically upward wi
Bumek [7]

Answer:

Both objects will undergo the same change in velocity

Explanation:

m = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of object

Any object which is falling has only the acceleration due to gravity.

ma=\dfrac{GMm}{r^2}\\\Rightarrow a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^6)^2}\\\Rightarrow a=9.81364\ m/s^2

The acceleration due to gravity on Earth is 9.81364 m/s²

So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.

Hence, both objects will undergo the same change in velocity.

4 0
3 years ago
A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
siniylev [52]

Answer:

m=1.53kg    

Explanation:

To solve this problem we use the Hooke's Law:

F=k*\Delta x     (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm  

The force applied is due to the weight

F=mg

We replace in (1):

mg=k*\Delta x  

We solve the equation for m:

m=k*\Delta x/g=5*3/9.81=1.53kg    

7 0
3 years ago
A combustion reaction usually gives off heat and light is it true or false
yanalaym [24]
The answer is true I think
3 0
3 years ago
Determine the length of the object shown <br> 1.2 cm <br> 1.3 cm <br> 1.25 cm <br> 1.250 cm
stiks02 [169]

The correct answer is 1.25 because it is 1/2 of 1 1/2 and that is 1.25.

6 0
2 years ago
Read 2 more answers
A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

\omega_i = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0
3 years ago
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