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ella [17]
3 years ago
10

You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim

e before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/ s 2.
How much distance is between you and the deer when you come to a stop?What is the maximum speed you could have and still not hit the deer?
Physics
1 answer:
andreev551 [17]3 years ago
5 0

Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

distance traveled in the reaction time d = v t

d = 21 * 0.50 = 10.5 m

deceleration after this time = -10 m/s^2

now the distance traveled by the car after applying bakes

v_f^2 - v_i^2 = 2a d

0 - 21^2 = 2(-10)d

d = 22.05 m

so total distance moved before it stop

d = 22.05 + 10.5 = 32.55 m

so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m

again by kinematics

v_f^2 - v_i^2 = 2 ad

0 - v^2 = 2(-10)(24.5)

v = 22.1 m/s

so maximum speed would be 22.1 m/s

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Answer:

During a typical school day all forms of eneergy is being utilised and also transfer of energy takes place from one form to another.

Explanation:

Chemical energy- A bunsen burner burning a beaker filled with water.

Heat energy- The water in the beaker absorbing the heat from the burner.

Electrical energy- Running Fans and lights in a classroom by switches.

Solar energy- Solar energy harnessed by solar panels to run the fans and lights by converting it into electrical energy.

Potential energy- A ball being held by a student at a certain height possesses energy due to gravity.

Kinetic energy- The same ball being left by the boy from a certain height produces kinetic energy

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A moving car has momentum. if it moves twice as fast its momentum is ____________ as much
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It's momentum is twice as much.
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A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
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Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

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3 0
2 years ago
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
3 years ago
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