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Drupady [299]
3 years ago
6

A car is going 3 m/s, then it speeds up for 4 seconds. If it has an acceleration of 5m/s^2, find its final velocity. Which equat

ion can you use?
A Δx = vi*t + (½)at^2
B vf = vi + a*t
C Δx = (½)(vf + vi)*t
D vf^2 = vi^2 + 2aΔX
Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0

Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.

Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !

vf = (initial velocity) + (acceleration)*(time)

vf = (3 m/s) + (5 m/s²)*(4 sec)

vf = (3 m/s) + (20 m/s)

vf = 23 m/s

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A proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
lawyer [7]

The change in potential energy of the proton is  5.6 x 10^{-17} Joule

<h3>What is a Uniform Electric Field ?</h3>

A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.

Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

  • The distance d = 10 cm = 0.1 m
  • Electric field E = 3.5 KN/C
  • Proton charge q = 1.6 x 10^{-19} C

The Work done = Fd

but F = Eq

Recall that Electric field E = F/q = V/d

Where V = potential difference.

Let us first calculate the V

E = V/d

V = Ed

Substitute all the parameters into the formula above

V = 3.5 × 10³ × 0.1

V = 350 v

from F/q = V/d

make F the subject of formula and substitute it in work formula

F = Vq/d

W.D = Vq/d x d

W.D = Vq

Substitute all the parameters into the formula above

W.D = 350 x 1.6 x 10^{-19}

W.D = 5.6 x 10^{-17} J

Work done = Energy = Potential Energy

Therefore, the change in potential energy of the proton is 5.6 x 10^{-17}<em> Joule</em>

<em />

Learn more about Electric Field here: brainly.com/question/14372859

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7 0
2 years ago
Identify each statement as an example of melting or sublimation,
xeze [42]

1st is sublimation

2and is melting

3red is melting

4th is sublimation

sublimation is just "skipping" the liquid phase / state

5 0
3 years ago
A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)
Contact [7]

You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

 y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

8 0
3 years ago
Read 2 more answers
The center of ?
maria [59]

Answer:

Gravity

Explanation:

Thats the answer bucko. Gives the brainliest or I die

6 0
3 years ago
An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

So we have v_{AT}=(95km/h)-(75Km/h)=20Km/h.

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s

And in that time the car would have traveled (<em>relative to the ground</em>):

d=v_At=(95Km/h)(0.065h)=6.175Km

If they are traveling in opposite directions, <u>we have to do all the same</u> but using v_{AT}=v_A+v_T (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

v_{AT}=(95km/h)+(75Km/h)=170Km/h

t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

5 0
3 years ago
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