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jeyben [28]
3 years ago
12

A motorcycle is traveling along a highway at 26 m/s. How far does the motorcycle travel in 15 s?​

Physics
1 answer:
leva [86]3 years ago
8 0
26 m/s for fifteen seconds. distance = rate times time, so distance = 26 m/s * 15 seconds. this gives you distance = 390 meters.
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Is it possible for a car to be accelerating to the west while it is driving to the east?
Ber [7]

Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

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What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?
vfiekz [6]

Answer: 0.0039\ A

Explanation:

Given

Diameter of the rod d=2.54\ cm

length of rod is l=20\ cm

Resistivity of silicon is \rho=6.4\times 10^2\ \Omega-m

cross-section of the rod A

\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2

Resistance of rod is  R

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\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega

Current is given by

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3 0
2 years ago
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5 0
3 years ago
Read 2 more answers
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T

the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

hence, the magnitude of the magnetic force is 0.187N

4 0
3 years ago
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