All categories

3 years ago

Two cars are moving with velocities 70km/hr and 50km/hr in east and west direction respectively.

Physics

1 answer:

viva [34]3 years ago

5 0

Answer:

70 + 50 = 120 km/hr

Explanation:

The driver of either car would see the other car approaching or departing at 120 kph

You might be interested in

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago

Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m

Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0

3 years ago

Read 2 more answers

two cars are traveling in opposite directions on the thruway. car a is traveling at 55mi/h and car b is traveling at 70 mi/h the

Tresset [83]

a minute or less than a minute

3 0

3 years ago

Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through

bekas [8.4K]

We know that arc length (x(t)) is given with the following formula:
$x(t)=\theta(t) r$
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
$r(\theta)=r_0-\frac{D\cdot \theta}{2\pi}$
When we plug this back into the first equation we get:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\$
We must solve this quadratic equation.
The final solution is:
$\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}$
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}$

8 0

4 years ago

Define output work and input work

Olenka [21]

Answer: Input work is the work done on a machine as the input force acts through the input distance.Other ways it would be the work done on a body or system, that is, forces that are applied to the body or system. This is in contrast to output work which is a force that is applied by the body or system to something else.Output work is the work done by a machine as the output force acts through the output distance. The machine does to the object to increase the output distance.

Explanation:

6 0

3 years ago