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iragen [17]
3 years ago
8

Calculate the momentum of a 953kg elephant running at a rate of 3.85 m/s.

Physics
1 answer:
cricket20 [7]3 years ago
6 0
Hi!
So we got 953kg = mass
And we also got 3.85 = speed
M/S = 953/3.85
953/3.85= 247.532468
ANSWER: 247.532468
Hope I helped! :D
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An automobile traveling 89.0 km/h has tires of 62.0 cm diameter. (a) What is the angular speed of the tires about their axles? (
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Answer:

a) 79.7rad/s

b) -18.7rad/s^2

c) 53m

Explanation:

We will use the MKS system of unit, so:

v=89.0km/h=89.0\frac{km}{h}*\frac{1000m.h}{3600km.s}=24.7m/s\\\\d=62.0cm=62.0cm*\frac{0.01m}{1cm}=0.62m

now, The angular speed is given by:

\omega=\dfrac{v}{\frac{d}{2}}\\\\\\\omega=\frac{24.7m/s}{0.31m}=79.7rad/s

in order to obtain the angular acceleration we have to apply the following formula:

(\omega_f)^2=(\omega_o)^2+2\alpha*\theta\\\\\alpha=-\frac{(\omega_o)^2}{2*rev*2\pi}\\\\\alpha=-\frac{(79.7m/s)^2}{2*27*2\pi}=-18.7rad/s^2

The linear displacement is given by:

d_l=\theta*r\\d_l=rev*2\pi*\frac{d}{2}\\\\d_l=27*2\pi*0.31m=53m

3 0
2 years ago
Your cat Goldie (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 39.930 m long and inclined 27.0 ∘ above t
Citrus2011 [14]

Answer:

The speed when se reaches the top of the incline is 0.28 m/s

Explanation:

The work done is equal to the change of kinetic energy, then:

Wg + Wf + Wn = ΔEk

Where

Wg = work done by gravity

Wf = work done by force

Wn = work done by normal force

-mgdsin\theta +Fd+0=\frac{1}{2} *m*(v_{2} ^{2} -v_{f} )

Where

m = 8.5 kg

g = 9.8 m/s²

d = 39.93 m

F = 37.4 N

vf = 2 m/s

Replacing:

39.93*(-8.5*9.8*sin27+34.4)=\frac{1}{2} *8.5*(v_{2} ^{2} -2^{2} )\\v_{2} =0.28m/s

7 0
3 years ago
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Answer:

vbmdgj

Explanation:

4 0
3 years ago
A 18 kg sled starts up a 28 degree incline with a speed of 2.3 m/s. The coefficient of kinetic friction between the sled and the
sergey [27]

Answer:

d = 0.391 m

Explanation:

given,

mass of sled = 18 kg

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kinetic friction between the sled and inclined = 0.25

using energy equation

\dfrac{1}{2}mv^2- Fd - mgh = 0

\dfrac{1}{2}mv^2- (\mu mg cos\theta )d - mgdsin\theta = 0

\dfrac{1}{2}mv^2 = (\mu mg cos\theta )d + mgdsin\theta

\dfrac{1}{2}mv^2 = d mg (\mu cos\theta+sin\theta)

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d = \dfrac{2.3^2}{2\times 9.8 (0.25\times cos28^0+sin28^0)}

d = 0.391 m

hence, the distance traveled by the sled is equal to 0.391 m

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3 years ago
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The answer & explanation for this question is given in the attachment below.

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