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3 years ago

What would happen if all the toilets were flushed at once?

Physics

2 answers:

irina1246 [14]3 years ago

6 0

Answer: sewer systems across the world would be overwhelmed

Explanation:

Sliva [168]3 years ago

5 0

Answer:

nothing would happen

Explanation:

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A tennis player hits a ball 2.0 m above the ground.

tangare [24]

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

As it hits the ground in time t, so put y = 0

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

your welcome, and have a great day.

8 0

2 years ago

Read 2 more answers

Cathy is moving into a tenth-floor apartment. She has a grand piano that weighs 570 pounds and is too big to fit in the elevator

Kazeer [188]

She will need pulleys. Nothing else will work

4 0

3 years ago

A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed

AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

$\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s$

And the frequency of the motion instead is given by:

$f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz$

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

$F=kA$

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

$F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N$

k = 160 N/m is the spring constant

Solving for A, we find

$A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m$

3 0

2 years ago

A 2 kg marble moving at 4 mi./s collides into a 1 kg marble at rest. After collision, the 2 kg marble speed decreased to 2 mi./s

Klio2033 [76]

Answer:

$2\sqrt{6}$ $\frac{mi}{s}$

Explanation:

Assuming there is no waste of energy:

$K_{1} = K_{2}\\\frac{1}{2}m_{1}v_{1_{1}}^{2} + \frac{1}{2}m_{2}v_{2_{1}}^2 = \frac{1}{2}m_{1}v_{1_{2}}^{2} + \frac{1}{2}m_{2}v_{2_{2}}^2\\\\=> m_{1}v_{1_{1}}^{2} + m_{2}v_{2_{1}}^2 = m_{1}v_{1_{2}}^{2} + m_{2}v_{2_{2}}^2\\\\m_{1} = 2 kg, m_{2} = 1 kg, v_{1_{1}} = 4 \frac{mi}{s} , v_{2_{1}} = 0\\=> 32 = 8 + v_{2_{2}}^{2} => v_{2_{2}} = 2\sqrt{6} \frac{mi}{s}$

6 0

2 years ago

Whic ones are true and false?

Tju [1.3M]

Answer:

I think option a is true

6 0

2 years ago

Read 2 more answers