All categories

2 years ago

A bus hits a bug and the bug splatters on the windshield, which force is greater?

Physics

1 answer:

grandymaker [24]2 years ago

8 0

The forces acting on a body and the type of motion that results are given by Newton's three Laws of motion

The force of the bus is the same as reactive force of the bug

Reason:

According to Newton's third Law of motion, given that the bug collides

with the bus, the force with which the bus hits the bug is equal to the

reactive force of the bug on the on the bus

According to Newton's second Law of motion, force is equal to the rate of

change of the momentum produced

The impulse of the force of the bus on the bug is given as follows;

F·Δt = m·(v₂ - v₁)

Given that the force of the bus is large, the change in momentum, m·(v₂ - v₁),

is also large such that the parts of the bug are split by the rapid change in

velocity, and the bug splatters on the windshield, and is then carried along

on the trip,

The equally large reactive force of the bug, is such that the bug splatters

due its magnitude

Therefore, the correct response is that the forces are the same

Lean more here:

brainly.com/question/21279060

You might be interested in

If a negatively charged particle enters a region of uniform magnetic field which is perpendicular to the particle's velocity, wi

lys-0071 [83]

Answer:

Same

Explanation:

While moving through a magnetic field in a direction perpendicular to a B-field, a continuous force experienced by a charged particle. If this magnetic field remains uniform, the force exerted also remains same and hence the velocity with which the particle is moving remains same. However, the particle is forced to move on a curved path until it forms a complete circle.

Hence, the kinetic energy remains the same because the speed is same

3 0

2 years ago

What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of

Dvinal [7]

Answer:1384 Hz

Explanation:

Given

wavelength $(\lambda )$ =0.25 m

Temperature T $=25^{\circ}$

at $T=25^{\circ}$ velocity of sound is 346 m/s

and we know

$velocity=frequency\times \lambda$

$v=f\times \lambda$

$346=f\times 0.25$

f=1384 Hz

5 0

3 years ago

Read 2 more answers

The sunlight hits the surface of a lake this will like caused

ziro4ka [17]

A). No. Condensation happens when you take heat out of a gas.

b). No. I'm not sure what transpiration is.

c). Yes. Evaporation happens when you add heat to a liquid.

d). No. Sublimation sometimes happens when you add heat to a solid.

3 0

3 years ago

Read 2 more answers

What is empirical evidence and what's is it not

ladessa [460]

Empirical evidence is information acquired by observation or experimentation. This data is recorded and analyzed by scientists and is a central process as part of the scientific method.
As example of not empirical evidence:
Language, letters, words, truths, numbers, logic, mathematics, on and on.
All of these things do not have empirical existence. They exist only in the mind.

3 0

2 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago