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2 years ago

A 30kg mass is brought from the earth to the moon what is the weight on the earth

Physics

1 answer:

ozzi2 years ago

8 0

Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N

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A 0.050 kg toy truck moving right at 0.20 m/s collided with a toy car weighing 0.015 kg initially at rest, on a frictionless tra

Kamila [148]

Answer:

0.1667 m/s

Explanation:

m1V1 + m2V2 = m1V3 + m2V4

0.01 = ( 0.0075) + (0.015 * V4)

V4 = (0.01 - 0.0075) / (0.015)

V4= 0.1667

7 0

3 years ago

Read 2 more answers

The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open squa

hichkok12 [17]

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

4 0

3 years ago

If an object's kinetic energy is zero, what is its momentum?

hammer [34]

If the object's kinetic energy is zero, then due to in multiplication factor, it's momentum will also be equal to zero 'cause the velocity of the object must be Nil

In short, Your Answer would be: "Zero"

Hope this helps!

6 0

3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

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3 years ago

Where does most star formation occur in the milky way galaxy?

k0ka [10]

Answer:

Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest.

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2 years ago