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Ilia_Sergeevich [38]
3 years ago
10

A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca

r 3.0 seconds to stop. What is the force required to stop the car?
A. 2.2 × 104 newtons
B. 2.5 × 103 newtons
C. 7.3 × 103 newtons
D. 1.4 × 103 newtons
Physics
3 answers:
Mariulka [41]3 years ago
4 0

i believe it is C....tell me if im right please<3

boyakko [2]3 years ago
4 0

Answer:

Force, F=7.3\times 10^3\ N

Explanation:

Given that,

Mass of the car, m = 1000 kg

Initial speed of the car, u = 22 m/s

Finally, the brakes are applied, v = 0

Time, t = 3 s

We need to find the force required to stop the car. It is given by using second law of motion as :

F=m\times a

F=\dfrac{m(v-u)}{t}

F=\dfrac{1000(22)}{3}

F = 7333.33 N

or

F=7.3\times 10^3\ N

So, the force required to stop the car is 7.3\times 10^3\ N. Hence, this is the required solution.

himiko2 years ago
0 0

yes it is c

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5 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

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3 years ago
A scientist that applies the laws of science to the needs of communities is called _____.
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Answer:

The experimental scientist

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2 years ago
From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

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8 0
3 years ago
Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

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radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}

m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

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hence no of bright fringe

M = m + 1

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M = 222 fringes

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3 years ago
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