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3 years ago

A car drives at steady speed around a perfectly circular track.

Physics

1 answer:

gayaneshka [121]3 years ago

6 0

Answer:

e. Both the acceleration and net force on the car point inward.

Explanation:

If no net force acts on the car, the car must drive in a straight line, at constant speed.

As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.

In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.

The net force that causes this acceleration, aims inward, and is called the centripetal force.

It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.

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Which type of wave results from a moderately sloping coastal region?

Naddik [55]

Hey there!

Your answer: Spilling breaker

Spilling breaker usually occurs when a beach or ocean is flat, and as the waves of the wind continues to happen, slowly the region would eventually become a slope.

It's almost like play-dough. Let's say that we set a perfect flat surface of play-dough on the table. As we continue slide our hands one direction, doesn't the play dough have more on one side than the other? It eventually contains a slope if you add enough from the first place.

Your answer: Spilling breaker

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2 years ago

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What information should be posted on or near a dishwasher?

Murljashka [212]

A.. in case of any problems that may occur you would know what company to call

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2 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

2 years ago

Why do electrons move from the negative end of the tube to the positive end

defon

Due to attraction ... of opposite charges

3 0

2 years ago

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

2 years ago