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3 years ago

The mass of an object is 60kg on the surface of the earth what will be its weight on the surface of the moon

Physics

1 answer:

iris [78.8K]3 years ago

8 0

Answer:

Wm = 97.2 [N]

Explanation:

We must make it clear that mass and weight are two different terms, the mass is always preserved that is to say this will never vary regardless of the location of the object. While weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 10 [m/s²]

But in order to calculate the weight of the body on the moon, we must know the gravitational acceleration of the moon. Performing a search of this value on the internet, we find that the moon's gravity is.

gm = 1.62 [m/s²]

Wm = 60*1.62

Wm = 97.2 [N]

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A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

ivann1987 [24]

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

(m)child(d)child)=(m)rock(d)rock)

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

6 0

3 years ago

What happens when a particle of matter meets its corresponding antiparticle of antimatter?

shutvik [7]

Answer:

The combined mass of the two particles is completely transformed into energy (photons). This process is called matter-antimatter annihilation.

Explanation:

6 0

3 years ago

A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T.

Degger [83]

Answer:

The angle the wire make with respect to the magnetic field is 30°

Explanation:

It is given that,

Length of straight wire, L = 0.6 m

Current carrying by the wire, I = 2 A

Magnetic field, B = 0.3 T

Force experienced by the wire, F = 0.18 N

Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :

$F=ILB\ sin\theta$

$\theta=sin^{-1}(\dfrac{F}{ILB})$

$\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\times 0.3\ T})$

$\theta=30^{\circ}$

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.

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3 years ago

Which of the filling is a result of the transfer of energy?

bija089 [108]

Explanation:

the speed of sound depends on the particles of the medium of the sound wave is moving through because the particles in the solid and tightly packed together South Wales weekly transfers energy from particle to particle in a solid

6 0

3 years ago

The smallest angle of inclined desk at which a book begins to slide off the desk is 15∘ relative to the horizontal.

Ray Of Light [21]

The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.

Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)

N = the normal reaction, which is equal to
N = mg cos(12°)

R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)

F = the component of the weight acting down the incline. It is
F = mg sin(12°)

Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
$\frac{sin(12^{o})}{cos(12^{o})} = \frac{\mu mg}{mg} = \mu \\\\
tan(12^{o}) = \mu$

Therefore, the static coefficient of friction is
μ = tan(12) = 0.213

Answer: μ = 0.21 (nearest tenth)

6 0

3 years ago

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