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Andrej [43]
3 years ago
13

A rock is thrown from a cliff and hits the ground five seconds later at a distance of 50 m from the cliff how high was a cliff

Physics
1 answer:
Iteru [2.4K]3 years ago
3 0

Answer:

122.5m

Explanation:

d= 1/2 (9.8) (5) ^2

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A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 5.39
kodGreya [7K]
<span>To do this question, we need to know that momentum is conserved, meaning the overall velocity of the two balls has to be the same before and after the collision.  </span>

<span>After collision... </span>

<span>Ball 1: 4.33m/s *cos 30 = 3.75 m/s (x-component) </span>
<span>4.33m/s * sin 30 = 2.165 m/s ( y-component) </span>

<span>Ball 2 (struck ball): 5 m/s - 3.75m/s = 1.25 m/s (x-component)  </span>
<span>-2.165 m/s (y-component) note: it has to be in the opposite direction to conserve momentum </span>

<span>tan-1(2.165/1.25) = 60 degrees </span>
<span>Struck ball's velocity = sqrt(1.25^2 + 2.165^2) = 2.5 m/s at 60 degree with respect to the original line of motion.  </span>

<span>Hope you understand!</span>
5 0
4 years ago
Read 2 more answers
Round 61.062 to one decimal place.​
RoseWind [281]

Answer: 61.1 is the answer I wish this answer help you.

3 0
4 years ago
Identify the energy transformation that takes when place when you apply the brakes on a bicycle.
valina [46]
Um Kinetic, mechanical, potential idk

The rider is riding the nick so kinetic then mechanical making the bike move then potential stopping the bike
4 0
3 years ago
A 0.30 kg yo-yo consists of two solid disks of radius 5.10 cm joined together by a massless rod of radius 1.0 cm and a string wr
docker41 [41]

Answer:

0.70046 m/s²

2.732862 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of yo-yo = 0.3 kg

R = Radius of rolling disk = 5.1 cm

r = Radius of rod = 1 cm

For a rolling disks the acceleration is given by

a=\frac{g}{1+\frac{R^2}{2r}}\\\Rightarrow a=\frac{9.81}{1+\frac{0.051^2}{2\times 0.01^2}}\\\Rightarrow a=0.70046\ m/s^2

The acceleration of the yo-yo is 0.70046 m/s²

The tension in the string will be

T=m(g-a)\\\Rightarrow T=0.3\times (9.81-0.70046)\\\Rightarrow T=2.732862\ N

The tension in the string is 2.732862 N

8 0
3 years ago
Una tubería con secciones circulares se emplea para el traslado de agua entre el tanque central y un tanque auxiliar. El tanque
charle [14.2K]

A) 20 L/s

B) 2.55 m/s

C) 10.20 m/s

D) 400.8 kPa

Explanation:

a)

In this problem, we know that the volume of the tank:

V=72 m^3  

is filled in a time of

t = 1 h

We also know that the volume flow rate of water in the pipe is constant in every point of the pipe, so it can be calculated directly from the two data above.

The volume of the tank, in liter, is

V=72 m^3 = 72,000 L

While the time, in seconds, is

t=1 h = 3600 s

Therefore, the volume flow rate in Liters per second is:

Q=\frac{V}{t}=\frac{72,000}{3600}=20 L/s

b)

The volume flow rate of water through the pipe can be also written as

Q=Av

where

A is the cross-sectional area of the pipe

v is the speed of the water

In section B, we have:

r = 0.050 m is the radius of section B

so, the cross-sectional area of section B is:

A=\pi r^2 = \pi (0.050)^2=0.00785 m^2

The volume flow rate in SI units is

Q=\frac{V}{t}=\frac{72.0m^3}{3600 s}=0.02 m^3/s

Therefore, the speed of the water in section B is:

v=\frac{Q}{A}=\frac{0.02}{0.00785}=2.55 m/s

c)

As in part B), we know that the volume flow rate must remain constant through the entire pipe.

So, the volume flow rate in section A of the pipe is still

Q=0.02 m^3/s

The radius of the pipe in section A is

r=0.025 m

Therefore, the cross-sectional area in section A of the pipe is

A=\pi r^2 = \pi (0.025)^2=0.00196 m^2

So, since we have

Q=Av

we can find the speed of water in section A:

v=\frac{Q}{A}=\frac{0.02}{0.00196}=10.20 m/s

d)

Here we want to find the gauge pressure in section B.

We know that:

p_A = 2.0 atm = 105,000 Pa is the pressure in section A

h_A=15.0m is the altitude of section A

v_A=10.20 m/s is the speed of water in section A

v_B=2.55 m/s is the speed of water in section B

We can write Bernoulli's equation:

p_A + \rho g h_A + \frac{1}{2}\rho v_A^2 = p_B + \frac{1}{2}\rho v_B^2

where

\rho=1000 kg/m^3 is the water density

p_B is the pressure in section B

And solving for pB, we find:

p_B = p_A + \rho g h_A + \frac{1}{2}\rho (v_A^2-v_B^2)=\\=205,000+(1000)(9.8)(15.0)+\frac{1}{2}(1000)(10.20^2-2.55^2)=400,769 Pa

Which is

p_B = 400.8 kPa

5 0
3 years ago
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