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2 years ago

6

A 250 g beach ball rolls across the sand with a speed of 11.16 km/h. First convert units to kg and m/ then determine the momentu

m of the ball

1 answer:

yuradex [85]2 years ago

6 0

Answer:

0.775 kg-m/s

Explanation:

Convert the units to the right unit forms necessary

250 g -> 0.25 kg

11.16 km/h -> 3.1 m/s

Now use the formula:

velocity

mass /

momentum / /

\ / /

\ / /

p = mv

p = 0.25 × 3.1 = 0.775 kg-m/s

Hope this helps you!

Bye!

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You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have

pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0

2 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

3 years ago

A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?

Olenka [21]

car starts from rest

$v_i = 0$

final speed attained by the car is

$v_f = 5 m/s$

acceleration of the car will be

$a = 3.6 m/s^2$

now the time to reach this final speed will be

$t = \frac{v_f - v_i}{a}$

$t = \frac{5 - 0}{3.6}$

$t = 1.39 s$

so it required 1.39 s to reach this final speed

6 0

2 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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2 years ago

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sveta [45]

1. GPE - 40 * 2 * 10 = 800j

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2 years ago