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nikdorinn [45]
3 years ago
5

A car heading north collides at an intersection with a truck of the same mass as the car heading east. If they lock together and

travel at 28 m/s at 37° north of east just after the collision, how fast was the car initially traveling? Assume that any other unbalanced forces are negligible.
what is the correct answer
34 m/s
17 m/s
26 m/s
68 m/s
...?
Physics
1 answer:
stealth61 [152]3 years ago
3 0

This phenomenon is an inelastic collision. The problem is easier to solve because the masses of the colliding bodies are equal. By conservation of momentum, the final momentum of the bodies as one is equal to the VECTOR sum of the momentum of individual bodies. 

For this problem, the final velocity of the two cars is the sum of the initial velocities of each car. The velocity of the car is just the component of the velocity along the north direction.

Using trigonometry,
v1 = vf * sin(theta)
v1 = 28 * sin(37)

v1 = 16.9 m/s
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In this example we will analyze the forces acting on your body as you move in an elevator. Specifically, we will consider the ca
dexar [7]

Answer:

The reading of the scale during the acceleration is 446.94 N

Explanation:

Given;

the reading of the scale when the elevator is at rest = your weight, w = 600 N

downward acceleration the elevator, a = 2.5 m/s²

The reading of the scale can be found by applying Newton's second law of motion;

the reading of the scale  = net force acting on your body

R = mg + m(-a)

The negative sign indicates downward acceleration

R = m(g - a)

where;

R is the reading of the scale which is your apparent weight

m is the mass of your body

g is acceleration due to gravity, = 9.8 m/s²

m = w/g

m = 600 / 9.8

m = 61.225 kg

The reading of the scale is now calculated as;

R = m(g-a)

R = 61.225(9.8 - 2.5)

R = 446.94 N

Therefore, the reading of the scale during the acceleration is 446.94 N

4 0
2 years ago
What is a scientific theory
Tomtit [17]

Answer:

They arn't usually guesses, but they are well made theorys or explanations.  Its a well-substantiated explanation of facts that have been confirmed in expirements.

3 0
3 years ago
A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and
Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ <em>F</em> = <em>N</em> - <em>W</em> = 0

• net horizontal force:

∑ <em>F</em> = <em>Fs</em> = <em>m a</em>

where

<em>N</em> = magnitude of normal force

<em>W</em> = car's weight

<em>Fs</em> = mag. of static friction

<em>m</em> = car's mass

<em>a</em> = <em>v</em> ²/<em>R</em> = mag. of the centripetal acceleration

<em>v</em> = car's speed

<em>R</em> = radius of curve

Now,

• compute the car's weight:

<em>W</em> = <em>m g</em> = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

<em>N</em> = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

<em>Fs</em> = 0.5 <em>N</em> = 7350 N

• solve for the (maximum) acceleration:

7350 <em>N</em> = (1500 kg) <em>a</em>   →   <em>a</em> = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = <em>v</em> ²/ (35 m)   →   <em>v</em> ≈ 13 m/s

4 0
3 years ago
Two examples of hydraulic brake​
rjkz [21]

Answer:

1). Brake pedal or lever

2). A pushrod also called an actuating rod

3 0
2 years ago
Por favor, necesito ayuda urgente 4. vectores perpendiculares de 25 y 40 unidades cada uno. Hallar gráfica y numericamente el ve
Darina [25.2K]

Answer:

4)  R = 47.17 units , 5)  R= 10,29 unidades, 6)   R= 2994,4 km ,    θ = -33,7

Explanation:

Este es un ejercicio de adición de vectores

4) como los vectores son perpendiculares.

Para encontrar la resultante podemos usar el teorema de pitoras

        R =  √ a² + b²

        R = √√ ( 25² + 40²)

        R =  47,17 unidades

5)  Este caso como el angulo es diferente de 90 debemos usar la relación de pitoras completa

        R² =  a² + b² + 2 a b cos θ

donde el angulo es entre los vectores a y b

        R² = 12² + 16² – 2 12 16 cos 40

        R²= 400 – 294,16

       R= 10,29 unidades

6) En este caso los dos desplazamientos son perpendiculares, por lo cual Usamos el teorema de Pitágoras

          R = √ (2400² + 1600²)

          R= 2994,4 km

para el angulo de este desplazamiento usamos trigonometría

           tan θ = y/x

           θ = tan⁻¹ y/x

           θ = tan⁻¹ 1600/(-2400)

           θ = -33,7  

TRASLATE

This is a vector addition exercise

4) as the vectors are perpendicular.

To find the result we can use the Poreor theorem

        R = √ a² + b²

        R = √ (25² + 40²)

        R = 47.17 units

5) This case, as the angle is different from 90, we must use the complete ratio of the pitoras

        R² = a² + b² + 2 a b cos θ

where the angle is between vectors a and b

        R² = 12² + 16² - 2 12 16 cos 40

         R² = 400 - 294.16

       R = 10.29 units

6) In this case the two displacements are perpendicular, which is why we use the Pythagorean theorem

          R = √ (2400² + 1600²)

          R = 2,994.4 km

for the angle of this displacement we use trigonometry

           tan θ = y / x

           θ = tan⁻¹ y / x

           θ = tan⁻¹ 1600 / (- 2400)

           θ = -33.7

3 0
2 years ago
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