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Colt1911 [192]
3 years ago
15

Suppose you adjust your garden hose nozzle for a hard stream of water. you point the nozzle vertically upward at a height of 1.5

m above the ground. when you quickly turn off the nozzle, you hear the water striking the ground next to you for another 1.8 s .

Physics
1 answer:
Lisa [10]3 years ago
7 0
For problems especially pertaining motion, it is best to illustrate the problem to help you understand the problem. The picture I've attached is my illustration based on what I understood from the problem. Suppose the diamond in the picture is the nozzle. It is placed 1.5 m above the ground (bold horizontal line). The water coming out of the nozzle follows the direction of the arrows until it falls to the ground next to you holding the nozzle. When you turn it off, the water at the topmost part slowly comes back to the ground in 1.8 seconds. 

Unfortunately, you weren't able to complete the problem. However, I would make a smart guess. I think it is logical that the problem would ask how high did the water shoot upwards from the nozzle, denoted as x. In order to solve this, we use the equations for free-falling objects:

t = √2h/g
1.8 = √2h/9.81
h = 15.9 m

To find the height of the water from the nozzle, we subtract the total height to 1.5 m to determine x.

x = 15.9 - 1.5 = 14.4 m

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Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
4 years ago
A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons
Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

5 0
3 years ago
Kepler's laws and explanations​
IgorC [24]

Answer:

<em>The first law states that</em> every planet describes an elliptical path about the sun as a single focus.

<em>The</em><em> </em><em>second</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The line joining the planet to the sun sweeps out equal areas in equal time intervals.

<em>The</em><em> </em><em>third</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The squares of the period of revolution is proportional to the cubes of the mean distance between the planet and the sun

5 0
3 years ago
Read 2 more answers
What is the frequency of a wave that has a wavelength of 0.39 m and a speed
gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

f =  \frac{c}{ \lambda}  \\

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

f =  \frac{86}{0.39}  \\  = 220.512820...

We have the final answer as

<h3>200 Hz</h3>

Hope this helps you

7 0
3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0
2 years ago
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