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Helga [31]
3 years ago
6

A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v

elocity of the cart to increase to 30 m/s to the right.
Physics
1 answer:
hram777 [196]3 years ago
3 0

Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

v is the final velocity = 30m/s

u is the initial velocity = 20m/s

t is the time = 5secs

m = 300kg

Get the net force:

Recall that: F = m(v-u)/t

F  = 300(30-20)/5

F = 60(30-20)

F = 60(10)

<em>F = 600N</em>

<em>Hence the net force acting on the car is 600N.</em>

<em></em>

<em></em>

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A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?
Andrew [12]

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
             (15 m/s) in the direction opposite to the direction you call positive
.

Displacement = (distance between start-point and end-point)
                           in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
                 = (1/2) (3 m/s²) (25 s²)  =  37.5 meters

Displacement =
                     37.5 meters in the direction opposite to the direction you call positive.

5 0
3 years ago
The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st
Rudiy27

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

a=\frac{\textup{Change in speed}}{\textup{Time}}

thus,

a=\frac{26.8224 - 0 }{2.70}

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a = 9.934 m/s²

6 0
2 years ago
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Komok [63]

Answer:

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Explanation:

6 0
2 years ago
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The work function for magnesium is 3.70 ev. what is its cutoff frequency?
alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

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7 0
1 year ago
Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding. Disregarding friction, what is the velo
lilavasa [31]

Answer:

The velocity of the Mr. miles is 17.14 m/s.

Explanation:

It is given that,

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m

We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,

v=\sqrt{2gh}

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v=\sqrt{2\times 9.8\times 15}

v = 17.14 m/s

So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.

5 0
3 years ago
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