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dmitriy555 [2]
3 years ago
15

How much work must be done to raise a 1100kg car 2m above the ground?​

Physics
2 answers:
N76 [4]3 years ago
6 0
21560J? i dont know
Xelga [282]3 years ago
4 0

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

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A mother pats a child every time he throws a chocolate wrapper in the dustbin. This is an example of (A) Observational Learning
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A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}

0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

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3 years ago
The first five questions refer to the following problem:
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8 0
3 years ago
Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in
olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

I=\Delta p = m\Delta v

where

m = 62.0 kg is the mass of the passenger

\Delta v is the change in velocity of the car (and the passenger), which is

\Delta v = 5.30 m/s - 0 = 5.30 m/s

So, the linear impulse experienced by the passenger is

I=(62.0 kg)(5.30 m/s)=328.6 kg m/s

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

I=F \Delta t

where in this case

I=328.6 kg m/s is the linear impulse

\Delta t = 0.812 s is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N

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