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baherus [9]
2 years ago
12

As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

of the black horse?
Physics
2 answers:
marishachu [46]2 years ago
8 0
2 min = 120 sec

120/15 = 8

The black horse represents 8 seconds.
Wewaii [24]2 years ago
5 0

Answer:

8 seconds

Explanation:

The time taken by the merry go round to complete one round is called time period.

Number of rounds = 15

Time taken = 2 minutes = 2 x 60 = 120 second

15 rounds are completetd in 120 seconds

1 round completed in 120 / 15 = 8 seconds

So, the time period is 8 seconds.

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As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i
Kitty [74]

Answer:

<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>

Explanation:

This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc

<em>ALSO</em><em>,</em><em> </em><em>PLEASE DO</em><em> </em><em>MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>

<em>Bonne</em><em> </em><em>journée</em><em> </em><em>;</em><em>)</em><em> </em>

6 0
1 year ago
Read 2 more answers
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
2 years ago
A cement factory emits 900 kilograms of CO2 to produce 1,000 kilograms of cement. A fully grown tree removes six kilograms of CO
Dmitriy789 [7]
<h3><u>Answer;</u></h3>

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

<h3><u>Explanation and solution;</u></h3>

From the information;

900 kg CO2 = 1000 kg Cement

1 tree = 6 kg CO2

1 acre = 200 trees

<em>100000 kg Cement will require;</em>

<em>=(900 × 100000)/1000</em>

<em>= 90,000 kg of CO2</em>

<em>But 1 tree = 6 kg of CO2</em>

<em>Number of trees = 90,000/6</em>

<em>                            = 15,000 trees </em>

<em>But, 1 acre = 200 trees</em>

<em>Number of acres = 15,000/200</em>

<em>                             = 75 acres of land </em>

Therefore;

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

6 0
2 years ago
Read 2 more answers
What is NOT one of the three primary resources that families have to reach financial goals?
Elodia [21]
What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
8 0
1 year ago
Hot combustion gases enter the nozzle of a turbojet engine at 260 kpa, 747oc, and 80 m/s. the gases exit at a pressure of 85 kpa
Aleksandr-060686 [28]

Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temp are to be determined.

 

 

Given:

 

T1 = 1020 K à h1 = 1068.89 kJ/kg, Pr1 = 123.4

P1 = 260 kPa

T1 = 747 degrees Celsius

V1 = 80 m/s ->nN = 92% -> P2 = 85 kPa

Solution:

From the isentropic relation,

Pr2<span> = (P2 / P1)PR1 = (85 kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>

 

There is only one inlet and one exit, and thus, m1 = m2 = m3. We take the nozzle as the system, which is a control volume since mass crosses the boundary.

 

h2a = 1068.89 kJ/kg – (((728.2 m/s)­2 – (80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) = 806.95 kJ/kg\

From the air table, we read T2a  = 786.3 K

5 0
1 year ago
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