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3 years ago

9

A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began

to accelerate?

1 answer:

Montano1993 [528]3 years ago

8 0

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 = 36.95 m/s

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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

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3 years ago

FM radio ________________. a. had a somewhat shorter range than AM radio, but better sound quality. b. was widely adopted in the

svetlana [45]

Answer:

(A) FM Radio had a somewhat shorter ranger than AM radio, but better sound quality.

Explanation:

FM Radio was invented in 1933 by Edwin Armstrong who was an American engineer. FM stands for frequency modulation and AM stands for Amplitude Modulation.

FM is used for most broadcasts of music and FM radio stations use a very high-frequency range of radio frequencies.

In FM Radio, the sound is transmitted through changes in frequency. Both FM and AM radio signals experience frequent change in amplitude, they are far less noticeable on FM.

When switching between stations, FM antenna is alternating between different frequencies, and not amplitudes and this produces a much clearer sound and allows for smoother transitions with little to no audible static.

FM signals can be interfered by barriers and this could affect the signal strength. FM Radio signals are more clearer in a mountainous area that has no barrier.

AM radio was able to carry signals farther than AM radio.

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3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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2 years ago

What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s

Ksivusya [100]

The net force is 270 N

Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 90.0 kg

$a=3.0 m/s^2$

Substituting, we find the net force on the object:

$F=(90.0)(3.0)=270 N$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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2 years ago

1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms.

lidiya [134]

Answer:

The answer would be 0.04ohms.

Explanation:

Hopefully this helps

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2 years ago

Read 2 more answers