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Stolb23 [73]
3 years ago
5

How much momentum will a dumb-bell of mass 10 kg transfer

Physics
1 answer:
frosja888 [35]3 years ago
6 0

We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:

Vf² = Vi² + 2ad

Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.

Given values:

Vi = 0m/s (dumbbell starts falling from rest)

a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)

d = 80×10⁻²m

Plug in the values and solve for Vf:

Vf² = 2(10)(80×10⁻²)

Vf = ±4m/s

Reject the negative root.

Vf = 4m/s

The momentum of the dumbbell is given by:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

m = 10kg

v = 4m/s (from previous calculation)

Plug in the values and solve for p:

p = 10(4)

p = 40kg×m/s

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sergeinik [125]

Answer:

6.86 * 10^8 m

Explanation:

Parameters given:

Mass of hot gas, m = 2 kg

Gravitational Force, F = 618.2 N

Mass of Alpha Centauri, M = 2.178 * 10^30 kg

The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:

F = (G*M*m) / r²

Where G = gravitational constant

Therefore,

618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²

=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2

r² = 4.699 * 10^17 m²

=> r = 6.86 * 10^8 m

We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.

The mean radius of Alpha Centauri is 6.86 * 10^8 m.

5 0
3 years ago
Read 2 more answers
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

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