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Citrus2011 [14]
2 years ago
10

A car traveling at 100 km/hr strikes and unfortunate bug and splatters it.the net force of impact is

Physics
1 answer:
Paladinen [302]2 years ago
7 0

Answer:

it's zero

Explanation:

it is there is your answer

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An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai
rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion:        Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

                       Vf = Vi + a×t

                       Vf = 0 + 2.4×3.6

                       Vf = 2.4×3.6

                       Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0
3 years ago
Hull (1943) had rats push a lever that required 21 grams of force to budge. After they had learned to push the lever in order to
Zina [86]

Years of research have demonstrated that rats are intelligent creatures who experience pain and pleasure, care about one another, are able to read the emotions of others, and would assist other rats, even at their own expense.

<h3>Experiments:</h3>

In trials carried out at Brown University in the 1950s, rats were trained to press a lever for food, but they stopped pressing the lever when they noticed that with each press, a rat in an adjacent cage would scream in pain (after experiencing an electric shock).

Rats were trained to press a lever to lower a block that was hanging from a hoist by electric shocks administered by experimenters. A rat was subsequently hoisted into a harness by the experimenters, and according to their notes, "This animal normally shrieked and wriggled sufficiently while dangling, and if it did not, it was jabbed with a sharp pencil until it exhibited indications of discomfort." Even if it wasn't in danger of receiving a shock, a rat watching the scenario from the floor would pull a lever to lower the hapless rodent to safety.

Learn more about experiments on rats here:

brainly.com/question/13625715

#SPJ4

6 0
2 years ago
Which ecosystem contains 32% of the world's producers
tangare [24]
The Answer Is D 32 % Of The worlds production is In the marine ecosystem
4 0
3 years ago
Which type radiation can be observed well from Earth's surface?
umka2103 [35]

Answer:

Eletromagnetic radiation which is also known as visible light.

Explanation:

4 0
3 years ago
Read 2 more answers
Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}

T_2=266.12K

Therefore, the final temperature of gas is 266.12 K

5 0
3 years ago
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