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xeze [42]
3 years ago
15

A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s^2. (a) How much time does it take for the beetle to reach

this speed? (b) A top-fuel dragster can go from 0 to 30 mi/h in 0.600 s. Find the acceleration (in m/s^2) of the dragster?
Physics
1 answer:
Firlakuza [10]3 years ago
8 0
Im going to tell you what to do but not the result. So pay close attention: the first thing you need to do is convert miles/h to m/s. Then for the part a) <span>divide the final velocity by the initial velocity. That will give you the amount of it will take to accelerate to the final velocity.Now for the part b you </span>use the formula v=vo+at. I hope this can help you
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An oceanic ridge might be compared to what continental landform? A) delta B) great plains C) mountain range Eliminate D) mountai
Pavlova-9 [17]
A mountain range because an ocean ridge is an underwater mountain hope this helps you 
6 0
3 years ago
Two plane mirrors are separated by 120°, as the drawing illustrates. If a ray strikes mirror M1 at a θ1 = 64° angle of incidence
tino4ka555 [31]

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

  = 90° - 64°

  = 26°

β + γ + 120° = 180°

γ = 180° - 120° - β

γ = 180° - 120° - 26°

γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

δ = 56°

According to the law of reflection,

angle of incidence = angle of reflection

θ2 = δ = 56°

Therefore, angle θ2 at which the light leaves mirror 2 is 56°

8 0
3 years ago
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next
zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

V=1.4*10^5m/s

Explanation:

From the question we are told that:

Electric field B=1.5*10N/C

Distance d=2 x 10^{-3}

At negative plate

Generally the equation for Velocity is mathematically given by

V^2=2as

Therefore

V^2=\frac{2*e_0E*d}{m}

V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}

V=\sqrt{19.2*10^9}

V=1.4*10^5m/s

5 0
3 years ago
I need it in the next hour or so!
PSYCHO15rus [73]

The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is

∑ F = F[a] - F[f] - F[air] = ma

3100 N - 200 N - F[air] = (650 kg) (3 m/s²)

Solve for F[air] :

F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)

F[air] = 3100 N - 200 N - 1950 N

F[air] = 950 N

3 0
2 years ago
Scientists have discovered convection currents inside Earth. Explain how these convection currents move and what layer(s) they o
fgiga [73]

Answer:

I don't know what to do

Explanation:

So

7 0
3 years ago
Read 2 more answers
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