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scoundrel [369]
2 years ago
9

:)

Physics
1 answer:
sweet-ann [11.9K]2 years ago
8 0

Answer:

D

Explanation:

1.) The reaction is at dynamic equilibrium.

A: Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

2.) Which statement about the reaction is necessarily correct?

A: Both calcium carbonate and sodium carbonate are being produced.

3.) Both calcium carbonate and sodium carbonate are being produced.

A: The reaction is reversible.

4.) What is the fastest motion that can be measured in any frame of reference?

A: 300,000 km/s

5.) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north.

Why are we able to use these motions to describe the motion relative to the ground?

A: The people are moving much slower than the speed of light so the ground acts as a frame of reference.

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Binding energy is the energy needed to
densk [106]

Answer:

Answer C

Explanation:

4 0
2 years ago
Solve this question with explanation <br> Best answer will be brainliest <br> 40 points given
Yuliya22 [10]

Answer:

12 km/h

Explanation:

Average Speed = Distance / Time (or rate)

Pick a point on the graph for Ian and plug in values.

For example, 20 minutes -> 4km

Hence, Average speed = 4km ÷ 20 minutes = 0.2 km/min

0.2 km/min × 60 = 12 km/h

7 0
3 years ago
Read 2 more answers
Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

7 0
3 years ago
A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat
Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

  • To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

So

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is \sqrt{3^2+(-5)^2} =\sqrt{9+25} = \sqrt{34}

3 0
3 years ago
As seen in the figure, a bullet with mass of 15.0-g is fired vertically and penetrates a block with mass of 2.5-kg and the block
rodikova [14]

Answer:

KE = 2.03 J

Explanation:

After impact, the kinetic energy of the bullet+block will convert to potential energy

½mv² = mgh

v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s

conservation of momentum during the collision

0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)

u = 16.4481 m/s

KE = ½mv² = ½(0.015)16.4481² = 2.0290499...

KE = 2.03 J

4 0
3 years ago
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