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astraxan [27]
3 years ago
11

Conservation of Momentum Practice

Physics
1 answer:
Yakvenalex [24]3 years ago
8 0

Answer:

Before: 0 m/s  

After: -4 m/s

Explanation:

Before: Since you and your beau started at rest, your beau initial velocity is 0 m/s.

After: Since we have to conserve momentum,

momentum before push = momentum after push.

The momentum before push = 0 (since you and your beau are at rest)

momentum after push = m₁v₁ + m₂v₂ were m₁ = your mass = 60 kg, v₁ = your velocity after push = 3 m/s, m₂ = beau's mass = 45 kg and v₂ = beau's velocity.

So, m₁v₁ + m₂v₂ = 0

m₁v₁ = -m₂v₂

v₂ = -m₁v₁/m₂ = -60 kg × 3 m/s ÷ 45 kg = -4 m/s

So beau moves with a velocity of 4 m/s in the opposite direction

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A river is flowing south at a rate of 3 m/s. Steven can roe directly across the river if he aims the raft 30 degrees. What rate
levacccp [35]

Answer:

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

Explanation:

Given data,

The river flowing south at the rate, v = 3 m/s

To reach the other side directly across the river, he aims the raft, Ф = 30°

The speed of his raft across the river is given by the formula,

                                          V = v / Sin Ф

                                             = 3 / Sin 30°

                                              = 6 m/s

Steven has to row at a speed to reach the same horizontal spot at the other side of the river is, V = 6 m/s

3 0
3 years ago
An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce
Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Frequency of pendulum is given by

f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

Given in the question

f_p=\dfrac{1}{2}f_s

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m

The unstretched length of the spring is 2.67 m

6 0
3 years ago
If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?
astraxan [27]

Mass of the object is given as

m = 1417 g = 1.417 kg

now the speed of object is given as

v = \frac{d}{t}

here we know that

d = 47 m

t = 90 s

now we will have

v = \frac{47}{90} = 0.52 m/s

now we will have kinetic energy of the object as

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(1.417)(0.52)^2

KE = 0.19 J

now the power is defined as rate of energy

so here we can find power as

P = \frac{KE}{t}

P = \frac{0.19}{90} = 2.14\times 10^{-3} W

so above is the power used for the object

3 0
3 years ago
Where does current flows maximum in? series connection or parallel connection?
notsponge [240]
Current flow depends on other things in addition to the circuit configuration. If the SAME voltage is applied to some arrangement of the SAME components, the greatest current will occur when they are all in parallel.
3 0
3 years ago
A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
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