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lara
8 days ago
9

A car traveling at 70 mph70 mph down the interstate collides with a bug trying to cross the highway. Which of the following stat

ements best describes this collision? The car exerts a smaller force on the bug than the bug exerts on the car. The car exerts a larger force on the bug than the bug exerts on the car. Neither exerts a force on the other. The bug gets smashed because it got in the way of the car. The car exerts a force on the bug but the bug does not exert a force on the car. The car exerts the same sized force on the bug as the bug exerts on the car.
Physics
1 answer:
Sav [3.1K]8 days ago
5 0
The force exerted by the car on the bug is identical to the force the bug applies back on the car.
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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
serg [3582]

Response:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.  

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we utilize the previously calculated work:

W_{ff}=-F_f*(hy/sin\theta)   Rearranging for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

Vf^2=Vo^2+2*a*d

Rearranging for 'a':

a=-3.844m/s^2

Now, using dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
1 month ago
A fan that is switched on for 1 minute uses 500 W usefully but also wastes 300 W through the emission of sound and heat. What's
Ostrovityanka [3204]

Explanation:

Efficiency can be expressed as the ratio of useful output to the total power consumed. E = \frac{useful}{total}

The fan delivers 500W as useful output while wasting 300W. Thus, the overall power consumed equals 800W (500 + 300).

6 0
24 days ago
A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.
Maru [3345]

Answer:

La magnitud del momento total es de 21.2 kg m/s y su dirección es de 39.5° respecto al eje x.

Explanation:

¡Hola!

El momento total se calcula como la suma de los momentos de las piezas.

El momento de cada pieza se calcula de la siguiente manera:

p = m · v

Donde:

p = momento.

m = masa.

v = velocidad.

El momento es un vector. La pieza de 200 g se mueve a lo largo del eje x, por lo que su momento será:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

La pieza de 300 g se mueve a lo largo del eje y. Su vector momento será:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

El momento total es la suma de cada momento:

Momento total = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Momento total = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Momento total = (16.4 kg m/s, 13.5 kg m/s)

La magnitud del momento total se calcula de la siguiente manera:

|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s

La dirección del vector de momento se calcula utilizando trigonometría:

cos θ = px/p

Donde px es el componente horizontal del momento total y p es la magnitud del momento total.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3 (39.5° si no redondeamos la magnitud del momento total)

<pFinalmente, la magnitud del momento total es 21.2 kg m/s y su dirección es 39.5° respecto al eje x.

6 0
1 month ago
For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8
ValentinkaMS [3465]

Answer:

Velocity = v = 2.24 m/s

Acceleration = a = 0.20 m/s²

Explanation:

Refer to the attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

8 0
1 month ago
A spaceship moves at a speed of 0.95 c away from the Earth. It shoots a star wars torpedo toward the Earthat a speed of 0.90 c r
Sav [3153]

Answer:

0.345 c

Explanation:

v_{se} = speed of spaceship with respect to earth = 0.95 c

v_{ts} = speed of torpedo with respect to spaceship = - 0.90 c

v_{te} = speed of torpedo with respect to earth

The velocity of the torpedo in relation to earth is calculated as

v_{te}=\frac{v_{se} + v_{ts}}{1 + \frac{v_{se}v_{ts}}{c^{2}}}

v_{te}=\frac{0.95c + (- 0.90 c)}{1 + \frac{(0.95 c)(- 0.90 c))}{c^{2}}}

v_{te} = 0.345 c

5 0
13 days ago
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