All categories

3 years ago

Water enters a shower head

Physics

1 answer:

zalisa [80]3 years ago

4 0

Answer:

13.1

Explanation:

thats what i put in for acellus and its right

You might be interested in

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

ANEK [815]

Answer: $\dfrac{\pi}{2} ms^{-1}$

Explanation:

Let $T$ be the time required to make one revolution.

Let $r$ be the radius of the circular path.

Let $d$ be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So, $d=2\pi r$

Given, $d=0.5m\\T=2sec$

$d=2\times \pi \times 0.5=\pi m$

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let $s$ be the speed of the ball.

$s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}$

So,the speed of the ball is $\frac{\pi }{2}ms^{-1}$

5 0

3 years ago

Did i draw the tangent line correctly?

Jet001 [13]

Answer:

yes i belevie so

Explanation:

8 0

3 years ago

The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the

Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

$s=ut+0.5at^2$

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, $a=-10.0 m/s^2$

$\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m$

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is $s=44.0 m$

The acceleration is same, $a=-10.0 m/s^2$

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

$v^2-u^2=2as$

$0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s$

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.

7 0

3 years ago

If a dog runs 25m east and then turns around and runs

Nostrana [21]

Answer:

The dog ran a total distance of 45m but he is only 5m away from the starting line

Explanation: When you add 25 to 20 you get 45 for the total distance and if he ran back in the same direction then you would subtract 20 from 25 and get 5m

8 0

3 years ago

What is the acceleration of a 10-N freely falling object with no air resistance?

ludmilkaskok [199]

About 9.81 m/s^2 . Hope that helps

5 0

3 years ago

Read 2 more answers