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Novay_Z [31]
3 years ago
11

G (where g=9.8 m/s2). If an object’s mass is m=10. kg, what is its weight?

Physics
1 answer:
Karolina [17]3 years ago
8 0

An object's weight is

           (its mass) x (the acceleration of gravity wherever the object is) .

The acceleration of gravity on or near the Earth's surface is about 9.8m/s².

So an object with 10kg of mass, as long as it's located on or near the
Earth's surface, weighs
                                       (10 kg) x (9.8 m/s²) = 98 kg-m/s² = <em>98 Newtons</em>.

                                                                      That's about 22.05 pounds.

If you take the object somewhere else, it's still 10 kg of mass, but
it weighs something different, depending on the gravity of the place
where you take it.


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Explanation:

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3 years ago
A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri
Drupady [299]

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

KE = I/2*\omega_o^2

The inertia is:

I=m/2*R^2=0.072kg.m^2

So, the kinetic energy will be:

KE = 888.26J

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf    where Kf=0;   Ko = 888.26J;  and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R   where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

-888.26=-0.96*π*N

N=294.5 turns

6 0
3 years ago
A car starts from rest and accelerates at a constant rate after the car has gone 50 m it has a speed of 21 m/s what is the accel
Leya [2.2K]

Answer:

4.41 m/s^2

Explanation:

(v_f)^2 - (v_i)^2 = 2a * change in distance

(21)^2 - (0)^2 = 2a * 50

a = (21^2)/(2*50)

a = 4.41 m/s^2

3 0
3 years ago
What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing a=76m^2

We know that density of air d=1.29kg/m^3

Speed at top surface v_2=290m/sec and speed at bottom surface v_1=150m/sec

According to Bernoulli's principle force is given by

F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N

4 0
3 years ago
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