1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
BigorU [14]
3 years ago
14

Consider a high-speed bus colliding head-on with an innocent bug. The force of impact splatters the unfortunate bug over the win

dshield. According to Newton's 3rd Law, which is greater, the force on the bug or the force on the bus?
Physics
1 answer:
Alex17521 [72]3 years ago
3 0

Answer:

The force on the bug and the force on the bus are the same.

Explanation:

Newton's Third Law of Motion states that, "For every action, there is an equal and opposite reaction" or it can be restated that If a high-speed bus exerts a force on an innocent bug, then the innocent bug must exert a force of equal magnitude and opposite direction back on the high-speed bus.

Note: though the forces are equal but the effect are not, in this case the innocent bug was splattered but the bus was not destroyed.

You might be interested in
Can you hear it? In the cartoon space rocket, why do you think you would not be able to hear the whoosh of the rocket engine in
Brums [2.3K]

Answer:

Because space is a void with no air flow

Explanation:

7 0
3 years ago
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
3. Imagine a 10kg block moving with a speed of 20m/s<br> calculate the kinetic energy of this block
MatroZZZ [7]
The formula of the kinetic energy is:
E_{k}  =  \frac{m v^{2} }{2}
where m is a mass of the object, v is speed of the object at the moment of time. So we have:
E_{k}  =  \frac{10* 20^{2} }{2}  = 2000J
The answer is 2000 Joules.
6 0
3 years ago
Maggie's truck decelerates from 18.00 m/s to rest in 3.30 s. If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculat
Free_Kalibri [48]

Answer:

look at my Explanation

Explanation:

If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculate the magnitude of the net (unbalanced) force that can cause the acceleration.

8 0
3 years ago
Other questions:
  • A 58.72 kg person holding a steel ball stands motionless on a frozen lake.
    11·1 answer
  • Suppose the Opera station broadcasts at 90.5MHz and the Rock and Roll station broadcasts at 107.1MHz. 1. Which station's signal
    12·1 answer
  • Name two types of small bodies that are found on the Kuiper belt.
    9·1 answer
  • How do very high density objects appear in an ultrasound?
    6·1 answer
  • A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
    10·2 answers
  • What does the object on the screen below model?
    11·2 answers
  • a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha
    6·1 answer
  • NEED THE ANSWER PLZ<br><br>What are lanthanides?
    5·1 answer
  • Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
    15·1 answer
  • An object with a mass of 2.0 kg accelerates at 2.0 m/s^2 when an unknown force is applied to it. What is the amount of the force
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!