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Studentka2010 [4]
2 years ago
5

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 200 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Physics
1 answer:
Fofino [41]2 years ago
6 0

Answer:

Force will be acting southward and the magnitude of force will be 1000 N

Explanation:

given,

mass of car = 1000 Kg

initial speed of the car (u) = 20 m/s

final speed of the car (v) = 0 m/s

distance to stop the car = 200 m

using equation of motion

v² = u² + 2 a s

0 = 20² + 2 x a x 200

400 a = -400

a = -1 m/s²

Now, we know

Force = mass x acceleration

F = 1000 x -1

F = -1000 N

- ve sign of force represent force will be acting in the opposite direction of motion.

Force will be acting southward and the magnitude of force will be 1000 N

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schepotkina [342]
20.3 divided by 3.0 will get u velocity and v times 3.0s 
3 0
2 years ago
A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?
uranmaximum [27]
Simply subtract the two velocities and divide by 8.1,

\frac{0 - 35}{8.1} = - 4.32

~~

I hope that helps you out!!

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~Zoey
5 0
2 years ago
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You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init
Lady_Fox [76]

Answer:

T_{f} = 85.89 ° C

Explanation:

The linear thermal expansion process is given by

      ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

     ΔV = V β ΔT

Let's apply this equation to our case

     ΔV / V = ​​-0.507% = -0.507 10-2

     ΔT = (ΔV / V)  1 /β

     ΔT = -0.507 10⁻²  1 / 1.15 10⁻³

     ΔT = -4.409

     T_{f} –T₀ = 4,409

     T_{f} = T₀ - 4,409

     T_{f} = 90.3-4409

     T_{f} = 85.89 ° C

6 0
2 years ago
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11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
2 years ago
The table shows the decay in a 59 g sample of bismuth 212 overtime.
Vikki [24]

Answer: C

14.75g

Explanation:

Given that the half life time = 60.5s

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N = decayed mass

At time t = 0, No = 59g

At time t = 60.5s,

N = No/2 = 59/2

= 29.5g

At time t = 121

N = 29.5/2 = 14.75g

Therefore N = 14.75g

7 0
2 years ago
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