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kvasek [131]
3 years ago
10

7. Two bikes travelling in the same direction move at a speed of 30 km/hr. The bikes are separated by a distance of 5 km. What w

ould the speed of the car travelling in the opposite direction be if it meets these bikes at an interval of 4 minutes?
Physics
1 answer:
Fantom [35]3 years ago
6 0

Answer:

Explanation:

Call the bike on the right A

Call the bike on the left B

The car begins it's time when it passes A

4 minutes later, it passes B.

But B has moved in 4 minutes and that is the key to the problem.

How far has B moved.

t = 4 minutes = 4/60 hours = 1/15 of an hour.

d = ?

rate = 30 km / hr

d = r * t

d = 30 km/hr * 1/15 hours = 2 km

The distance between the bikes is 5 km.

So the car has traveled 5 - 2 = 3 km

d = 3 km

r = ?

t = 4 minutes = 1/15 hour

r = d/t = 3/(1/15)= 3 / 0.066666666  = 45 km/hr.

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The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

<h3>What is velocity?</h3>

Velocity is a vector quantity that tells the distance an object has traveled over a period of time.

Displacement is a vector quality showing total length of an area traveled by a particular object.

Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?

The slope of the line is equal to zero and the object will be stationary.

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1 year ago
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A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb
densk [106]
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  • Power=P=60W

  • Current=I

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2 years ago
How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

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Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

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q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

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Answer:

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