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3 years ago

11

a car of mass 1150 kg drives in a circle of radius 44 m. if the car has a speed of 13 m/s what is the centripetal force acting o

n the car?

1 answer:

Oksana_A [137]3 years ago

6 0

Answer:4417 N

Explanation:

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So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2.

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3 years ago

A coil consisting of 50 turns of wire is powered by a 9V battery. If the coil draws 1.5 A from the battery, what is the MMF, in

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A. 75 ampere-turns

<u>Explanation:</u>

Given:

Number of turns, n = 50

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MMF is independent on voltage and dependent on the number of turns and current in the wire.

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4 years ago

Read 2 more answers

The oil level in a tank is 2 m above the ground. The tank cover is air tight and the air pressure above the oil surface is 150 k

Andrew [12]

Answer:

a) 24.692 m/s

b) 19.4 m

Explanation:

To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:

$\frac{P1}{pg}+\frac{V1^2}{2g}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}+Z2$

We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:

$\frac{P1}{pg}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}$

a) The velocity (V2) is:

$\frac{P1}{pg}+Z1=\frac{V2^2}{2g}$

$(\frac{P1}{pg}+Z1)(2g)=V2^2$

$V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}$

Substituting the known values we can get the velocity at the out:

Atmospheric pressure= 101000 Pa

Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3

$V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}$

$V2=24.692 m/s$

b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:

$\frac{P2}{pg}+\frac{V2^2}{2g}+Z2=\frac{P3}{pg}+\frac{V3^2}{2g}+Z3$

We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:

$\frac{V2^2}{2g}=\frac{P3}{pg}+Z3$

$Z3=\frac{V2^2}{2g}-\frac{P3}{pg}$

Substituting the known values, the height (Z3) is:

$Z3=\frac{(24.692 m/s)^2}{2(9.81 m/s2)}-\frac{101000 Pa}{(9.81 m/s)(880 kg/m3)}$

Z3=Maximum Height=19.376=19.4 m

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4 years ago