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OverLord2011 [107]
3 years ago
15

What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s

Physics
1 answer:
nika2105 [10]3 years ago
7 0

Answer:

<em>An average net force of 5200 N is needed to stop the car</em>

Explanation:

<u>Cinematics and Dynamics</u>

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

\displaystyle a=\frac{v_f-v_o}{t}

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

\displaystyle a=\frac{0-28}{10.5}

a = -2.67~m/s^2

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

F = m.a =1950*2.67

F = 5200 N

An average net force of 5200 N is needed to stop the car

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7 0
3 years ago
A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th
Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

now we have

W_{gravity} = -mg(Lsin\theta)

W_{friction} = -\mu mgcos(\theta) L

so from above equation

Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

so from above equation work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

8 0
3 years ago
If a train is travelling 200km/hour eastward for 1800 seconds how far does it travel?
Olenka [21]

Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

v=\dfrac{d}{t}

d=v\times t

d=55.55\times 1800

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

5 0
3 years ago
Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 9.5 cm from the center of the bul
MA_775_DIABLO [31]

Given Information:  

Power of bulb = w = 25 W atts

distance = d = 9.5 cm = 0.095 m

Required Information:  

Radiation Pressure = ?

Answer:

Radiation Pressure =7.34x10⁻⁷ N/m²

Explanation:

We know that radiation pressure is given by

P = I/c

Where I is the intensity of radiation and is given by

I = w/4πd²

Where w is the power of the bulb in watts and d is the distance from the center of the bulb.

So the radiation pressure becomes

P = w/c4πd²

Where c = 3x10⁸ m/s is the speed of light

P = 25/(3x10⁸*4*π*0.095²)

P = 7.34x10⁻⁷ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 9.5 cm from the center of the bulb is 7.34x10⁻⁷ N/m²

4 0
3 years ago
The rock cycle _____.
Reika [66]
B is the answer you need and i honestly got this question on a middle school test

you must be in different area then me

4 0
3 years ago
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