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Ber [7]
4 years ago
11

The source of energy that keeps the Sun shining today is _________.

Physics
1 answer:
Reil [10]4 years ago
6 0

The source of energy that keeps the Sun shining today is _________.

<u>Is nuclear fusion</u>

~Hope this answers your question!~

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If a cart goes around a turn at 20 km/h ,what remains constant
Vitek1552 [10]

Answer: 4.speed

Explanation:

In this case, we know that the cart remains at a constant 20km/h.

Now, one could say that "the velocity remains constant, because it always is 20km/h"

But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.

But the module of the velocity, the speed, remains constant at 20km/h.

Then the correct option is 4, speed.

7 0
3 years ago
What is the current in the 10.0 resistor?
Flura [38]

Answer:

12.0V

12.0V

Explanation:

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8 0
3 years ago
A bus initially moving at 20 m/s with an acceleration of -4m/s2 for 5
Alecsey [184]

Answer:

50m; 0m/s.

Explanation:

Given the following data;

Initial velocity = 20m/s

Acceleration, a = - 4m/s²

Time, t = 5secs

To find the displacement, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Substituting into the equation, we have;

S = 20*5 + \frac {1}{2}*(-4)*5^{2}

S = 100 + (-2)*25

S = 100 - 50

Displacement, S = 50m

Next, to find the final velocity, we would use the third equation of motion;

V^{2} = U^{2} + 2aS

Where;

  • V represents the final velocity measured in meter per seconds.
  • U represents the initial velocity measured in meter per seconds.
  • a represents acceleration measured in meters per seconds square.

<em>Substituting into the equation, we have;</em>

V^{2} = U^{2} + 2aS

V^{2} = 20^{2} + 2(-4)*50

V^{2} = 400 - 400

V^{2} = 0

Final velocity, V = 0m/s

<em>Therefore, the displacement of the bus is 50m and its final velocity is 0m/s.</em>

8 0
3 years ago
Someone please help me
vagabundo [1.1K]
11.11  degrees Celsius
6 0
3 years ago
Read 2 more answers
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
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