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UNO [17]
3 years ago
8

A car traveling at 26.8m/s hits the brakes to slow down before turning into a neighborhood. Determine how fast it is going after

slowing down at a rate of -8.2m/s2 for 14.3seconds.
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

The speed of the car after the break was applied is  90.46 m/s

Explanation:

Given;

initial velocity of the car, u = 26.8 m/s

accelertation of the car, a = -8.2 m/s²

time of motion, t = 14.3 s

final velocity, v = ?

Apply the following kinematic equation;

v = u + at

v = 26.8 + (-8.2 x 14.3)

v = 26.8 -  117.26

v = -90.46 m/s

magnitude of the final velocity = 90.46 m/s

Therefore, the speed of the car after the break was applied is  90.46 m/s

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Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10
daser333 [38]

3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

                        = 64998 rad/min

Formula used,

v=rw where,

v is the Linear velocity

r is the radius

w is the angular velocity

v = 6 cm × 64998rad/min

  = 3,89,988 cm/min

Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.

Learn more about Angular speed here brainly.com/question/540174

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8 0
2 years ago
Which adaptation is likely to increase the chances of survival of an animal in a rainforest?
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Black-spotted skin coat as camouflage while stalking prey.
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What do we call a substance in<br> which two or more elements are<br> chemically bonded
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A compound

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A 0.10 g honeybee acquires a charge of +23 pC while flying.
kari74 [83]

Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

Therefore the ratio :

\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

\frac{F}{w} =2.347\times 10^{-6}\ N

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
3 0
3 years ago
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