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3 years ago

Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum

Physics

2 answers:

lord [1]3 years ago

7 0

150kg ms is your answer

soldi70 [24.7K]3 years ago

3 0

We Know, P = m*v
Here, m = 30 Kg
v = 5 m/s

Substitute it into the expression,
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps!

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A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti

Alenkinab [10]

Answer:

$v_f = 3.45 m/s$

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

$mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 10 kg

$\theta = 10^0$

L = 3.5 m

so we will have

$10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)$

$2(9.81) sin10 \times 3.5 = v_f^2$

so final speed is given as

$v_f = 3.45 m/s$

6 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago

A parallel beam of light containing orange (610 nm) and violet (410 nm) wavelengths goes from fused quartz to water, striking th

Fudgin [204]

Answer:

0.98°

Explanation:

First we find the refracting angle of orange in water

Which is given as

စr= sin^-1. ( 1.456 sin60°/1.33)

= 71.2°

Then that of violet in water

စv=sin^-1. ( 1.468sin60°/1.342)

= 72.3°

So angle between boths colours is the difference

71.2- 72.3= 0.98°

6 0

4 years ago

In space, there is no friction. Without the force of friction, satellites and other moving objects will continue moving at the s

Yakvenalex [24]

Answer:

Newton's first law

Explanation:

it says that an object will continue to be at rest or keep moving till any external for is applied to it to make it move or stop respectively.

6 0

3 years ago

A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a cir

topjm [15]

Answer:

Explanation:

In this question it is assumed that expressions for speed and mass are asked

Kinetic energy

$KE= 0.5mv^2$

since the charge moving in a circle

$qvB= mv^2/r$

So we can write

$v= \frac{2KE}{qBr}$

so, the expression for speed

$v= \frac{2KE}{qBr}$

now to obtain the expression for mass

Force F = m×a

and so solving for m we get:

plug expression for v back into KE equation

$KE= \frac{1}{2}m(\frac{2KE}{qBr})^2$

now rearranging the above equation we get

$m= \frac{(qBr)^2}{2KE}$

5 0

3 years ago