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Tatiana [17]
3 years ago
8

A race car accelerates from 0.00m/s to 40.0m/s with a displacement of 50.0m. What was the cars

Physics
1 answer:
Vladimir79 [104]3 years ago
7 0

Answer:

Explanation:

Using the equation of motion v² = u²+2as

v is the final velocity = 40m/s

u is the iniyail velocity 0m/s

a is the acceleration

s is the displacement

Substituting in the formula;

40² = 0²+2a(50)

1600 = 100a

Divide both sides by 100

100a/100 = 1600/100

a = 16

Hence the car acceleration is 16m/s²

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Can someone help me on question 1?
grigory [225]
In question 1, both of your answers are correct, but I don't understand the process you went through in the 'a' part.

R = v/I . That's a correct formula.
But it doesn't help you in this form, because you need to find I
So turn it into a helpful form ... Solve it for I, so it says I=something.

R= v/I

Multiply each side by I : R I = V.

Now divide each side by R: I= V/R .
THERE'S the equation you want.

I = V / R

I = 1.5 / 10 = 0.15 Amp.

That's slightly cleaner, although I don't really understand what you were actually thinking in that part.

But again ... You answered both parts correctly, and your process in b is fine.
6 0
3 years ago
Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time
anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let \overset \rightarrow {v_y} represent the velocity of the boat across the river in the direction, \overset \rightarrow {d_y}, and let, \overset \rightarrow {v_x}, represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = \overset \rightarrow {v_y}

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is \overset \rightarrow {v_y}, such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

\overset \rightarrow {v_y} = 3 m/s

\overset \rightarrow {v_x} = 4 m/s

The width of the river, w_y = 6 meters, we have;

  • The resultant velocity = \sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s

The direction, θ, is given as follows;

\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}

The length of the path of the boat, <em>l</em>, is given as follows;

l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = \dfrac{l}{v}, therefore;

  • t = \dfrac{10}{5} = 2
  • The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

t = \dfrac{w_y}{v_y}

Therefore;

t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

brainly.com/question/15907242

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