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Harman [31]
4 years ago
14

A rock with a mass of 25 kg has a weight of 40.8 n on the moon. what is the value of the acceleration of gravity (g) on the moon

?
Physics
1 answer:
zaharov [31]4 years ago
7 0
Mass m = 25 Kg;  Force = Weight = mg   Force = ma  but a = g

          F = 40.8 N

Required: gravity g = ?

                F = mg

                g = F/m

                g = 40.8 N/25 Kg

                g = 1.6 m/s²
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A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl
Genrish500 [490]

Answer:

Explanation:

Given

mass of block m=5.7\ kg

at t=0 s

displacement is x=-0.7\ m

velocity v=-0.8\ m/s

acceleration a=2.7\ m/s^2

suppose x=A\cos (\omega t+\phi )   is the general equation of SHM

where A=amplitude

\omega=natural frequency of oscillation

therefore velocity and acceleration is given by

v=-A\omega \sin (\omega t+\phi )

a=A\omega ^2\cos (\omega t+\phi )

for t=0

-0.7=A\cos (\phi )---1

v=-0.8=-A\omega \sin(\phi)---2

a=2.7=-A\omega ^2\cos(\phi )----3

divide 1 and 3 we get

\omega ^2=\frac{27}{7}

\omega =\sqrt{\frac{27}{7}}

Now square and 1 and 2 we get

(0.7)^2+(\frac{0.8}{\omega })^2=A^2

A^2=0.49+0.166

A=0.81\ m

     

3 0
3 years ago
Why was John Paul Jones's victory important in the war? His victory increased France's confidence in the skill of the Continenta
IRINA_888 [86]

Answer:

C. His victory against a superior British foe inspired the American troops.

Explanation:

John Paul Jones is considered the hero in the Revolutionary War. He is known as the Father of the US Navy.

In the Revolutionary War, Jones sided with the American colonists against the British and took hold of naval ships. In 1779, when the British warship <em>Serapis</em> was in conflict with the American warship <em>Bon Homme Richard, </em>Jones plugged the American warship with the Britisher's warship and tossed a grenade into the opponent warship. Thus when Jones was victorious in the war, this boosted the American spirits for the war.

Therefore, option C is the correct answer.

4 0
3 years ago
Read 2 more answers
A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t
Gelneren [198K]

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04  x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

V=\dfrac{4}{3}\pi r^3

V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb  + N = m g

m=Mass of the ball = Density x volume

m = γ V    , γ =Density of the Ball

ρ V g  + N =  γ V g               ( take g= 10 m/s²)

\gamma =\dfrac{N+\rho V g}{V g}

\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times  10}{219.13\times 10}

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

5 0
3 years ago
How much heat is contained in 100 kg of water at 60.0 °C?
attashe74 [19]
First we need to write down heat capacity for water which is constant.
cp=4186 J/(kg*K)

The equation for Energy that we will be calculating is:

E=cp*m*T
where m is mass and T is absolute temperature (273,15 + 60 in this case). Replacing all the values in equation we get:

E = 4186*100*333,15 = 139 456 590 J
6 0
3 years ago
Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off?
Sunny_sXe [5.5K]

Answer:

Yes it is possible to control to some extent.

Explanation:

In general there are two types of magnets : permanent and temporary (electromagnets).

Electromagnets can be controlled since it basically depends on electricity. By switching on and off the electric supply the magnets also can be switched on and off respectively. We can also control the intensity of magnetic power.

On the other hand permanent magnet cannot be switched on and off but the magnetic properties can be altered event to an extent when it loses all its magnetic properties. It can be caused by high temperature, physical impact and also exposure to other magnetic fields. For every element there is a point of temperature called curie temperature above which the permanent magnet loses its magnetic properties. This can be brought back again by induced magnetism. The only issue is that induced magnetism work in most cases but not in all.

6 0
4 years ago
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