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dalvyx [7]
3 years ago
8

With increasing depth, the temperature within earth

Physics
1 answer:
hoa [83]3 years ago
4 0

Answer: im pretty sure it increases

Explanation:

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How do magnets work?
Ainat [17]
How do magnets workssss?
5 0
4 years ago
Read 2 more answers
A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system,
NARA [144]

As per the question the initial speed of the car [ u] is 42 m/s.

The car applied its brake and comes to rest after 5.5 second.

The final velocity [v] of the car will be zero.

From the equation of kinematics we know that

                                         v=u+at [ here a stands for acceleration]

                                         0=42 +5.5a

                                         a =\frac{-42}{5.5} m/s^2

                                          a= -7.64 m/s^2

Here a is taken negative as it the car is decelerating uniformly.

We are asked to calculate the stopping distance .

From equation of kinematics we know that

                                              S=ut+\frac{1}{2} at^2  [here S is the distance]

                                                      = 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m

                                                       =115.445 m      [ans]  

8 0
3 years ago
In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g
Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η=\frac{electric power}{water power}

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0
3 years ago
A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100
Verizon [17]

Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)  

4 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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