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3 years ago

(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.

Physics

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

16.7 s

Explanation:

T= Vf - Vo a= F

a m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s

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Apply Pascal's law to a ketchup packet. Explain what will happen if you tear a small

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Answer:

its will explode because it's like a soda with baking soda added together and you shake it the small hole can make a huge explosion

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3 years ago

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A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

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3 years ago

If the pressure on an ideal gas is increased, what will happen to the volume of the gas?

iogann1982 [59]

B). it will decrease.

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Hope this helps!

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3 years ago

A hollow, conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has a uniform surface charge density

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A. What is the new charge density

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2 years ago

A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .

xenn [34]

Answer:

The time taken by the bar to reach the bottom t=4.886s

Given:

Displacement of the bar S=9.2m

Angle of inclination $\theta=8.0^{\circ}$

Coefficient of friction factor $\mu k=0.056$

To find:

How long it takes to reach the bottom ‘t’

Step by Step Explanation:

Solution:

We know that the formula for weight of the soap bar is given as

$F_{g}=m g \sin \theta$

The frictional force acting on this soap bar is determined by

$F_{f}=\mu m g \cos \theta$

To determine the constant acceleration of the bar, we derive as

$F=m a$

Here $F=F_{g}-F_{f}$ and thus

$F_{g}-F_{f}=m a$ $m g \sin \theta-\mu m g \cos \theta=m a$

$a=g \sin \theta-\mu g \cos \theta$

Where $F_{g}$ =Force imparted due to weight

$F_{f}$ =Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

$\sin \theta$ and $\cos \theta$ are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

$s=\frac{1}{2} a t^{2}$ , From this

$a t^{2}=2 s$

$t^{2}=\frac{2 s}{a}$

$t=\sqrt{\frac{2 s}{a}}$

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

$t=\sqrt{\frac{2 \times 9.2}{a}}$ and we know that

$a=g \sin \theta-\mu g \cos \theta$

$=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0$

$=1.364-0.543$

$a=0.821$

$t=\sqrt{\frac{2 \times 9.2}{0.821}}$

$t=\sqrt{\frac{19.6}{0.821}}$

$t=\sqrt{23.87332}$

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

3 0

3 years ago