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lubasha [3.4K]
4 years ago
11

Arunner has a speed of 23 m/s. They see the finish line and speed up to 27 m/s. This happens in 5 seconds. If the runner has a m

ass of 78 kg, with what force did the
runner cross the finish line? Show all work to receive full credit.
Type your answer.
Physics
1 answer:
sp2606 [1]4 years ago
6 0
The answer is 100

Have a great day
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The variables for this experiment include mass, volume, and the materials in the various balls and their densities. In Part III,
zmey [24]

Answer:

volume and density

Explanation:

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8 0
2 years ago
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With what force will a car hit a tree if the car has a mass of 3,000 kg and an acceleration of 2m/s squared
julsineya [31]

Answer:

<h2>6000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3000 × 2

We have the final answer as

<h3>6000 N</h3>

Hope this helps you

7 0
3 years ago
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Una atracción de feria consiste en lanzar un trineo de 2 kg por una rampa ascendente que forma un ángulo de 30º con la horizonta
Arada [10]

Answer:

La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s

Explanation:

La masa del trineo, m = 2 kg

El ángulo de inclinación del trineo, θ = 30 °

El coeficiente de fricción, μ = 0,15

La altura a la que debe ascender el trineo, h = 4 m

La reacción normal del trineo en la rampa, N = m · g · cos (θ)

La fuerza de fricción F_f = N × μ

Dónde;

g = Th aceleración debida a la gravedad ≈ 9.81 m/s²

∴ F_f  = 0.15 × 2 kg × 9.81 m/s² × cos (30°) ≈ 2.55 N

La longitud de la rampa que se mueve el trineo, l = h/(sin(θ))

∴ l = 4/(sin(30°)) = 8

La longitud de la rampa que se mueve el trineo, l = 8 m

El trabajo realizado sobre la fricción, W_f = F_f × l

W_f = 2.55 × 8 ≈ 20.4

El trabajo realizado sobre la fricción, W_f ≈ 20.4 Julios

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. = m·g·h

∴ P.E. = 2 kg × 9.81 m/s² × 4 m ≈ 78.48 Joules

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules

El trabajo total requerido para levantar el trineo 4 metros por la rampa, W_t = W_f + P.E.

W_t = 20.4 J + 78.84 J ≈ 99.24 J

Energía cinética, K.E. = 1/2 × m × v²

La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;

K.E. = 1/2 × 2 kg × v² ≈ 99.24 J

∴ v² ≈ 99.24 J/(1/2 × 2 kg)

La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s

4 0
3 years ago
A blue car with mass mc = 427 kg is moving east with a speed of vc = 20 m/s and collides with a purple truck with mass mt = 1282
polet [3.4K]

Answer:8540 kg-g/s

Explanation:

Given

mass of blue car m_c=427 kg

velocity of blue car v_c=20 m/s

mass of the truck m_t=1282 kg

speed of truck v_t=13 m/s

After collision they stick and lock together

Let v be the velocity of combined system at angle \theta from vertical

Conserving momentum in east direction

m_c\times v_c=(m_c+m_t)v\cos \theta

427\times 20 =1709\times v\cos \theta------1

Conserving Momentum in Y direction

m_t\times v_t=(m_c+m_t)v\sin \theta

1282\times 13 =1709\times v\sin \theta-------2

squaring and then adding 1 & 2 we get

(8540)^2+(16666)^2=(1709)^2\cdot v^2

v=10.95 m/s

initial momentum of car=427\times 20=8540 kg-m/s

6 0
4 years ago
if a student wanted to design an experiment that uses sound scientific principles, which of the following should he or she not d
nikklg [1K]
Things he should do
1) Take several readings.
2) And find the average of the readings to get the accurate value.
Things should not do
1) Avoid human error.
2) Zero error
4 0
3 years ago
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