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trasher [3.6K]
3 years ago
10

Is it possible for an object to be going 100 miles per hour but have a velocity of zero?

Physics
1 answer:
Lesechka [4]3 years ago
5 0
It is possible for on object to be going at 100 miles per hour, but still have a velocity. This is because the object going at 100 miles per hour has speed, which is a scalar quantity, which is defined by only magnitude, but the velocity of the object can be 0, since velocity is a vector quantity which is defined by both magnitude and direction.

Since this object only has magnitude and no direction (which is not given), then the velocity can be 0
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What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?
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E=F*d/2 = M*g*d/2 = 25kg * 9.8 N/kg *0.65m / 2 = 79.625J
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A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
Masja [62]

Answers is  F=7.84 N

Friction force resists the effect of horizontal force and trying to approch to a limiting force.

we have formula for limiting friction force between block and floor

               F=Ц N

where N=mg

putting values we get answer.

Download pdf
3 0
2 years ago
A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated
Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) C=16.7pF

(b) r_{a} =3.749cm

(c) E=2.24*10^{4} N/C

Explanation:

Given data

Q=3.50nC\\V=210V\\r_{b}=5.0cm

For part (a)

The Capacitance given by:

C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF

For part (b)

The Capacitance of coordinates is given as

C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm

For part (c)

The electric field according to Gauss Law is given by:

EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2}  }=\frac{kQ}{r_{a}^{2}}\\  E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C

7 0
3 years ago
Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat
kvasek [131]

Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

8 0
3 years ago
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