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marysya [2.9K]
2 years ago
9

Your older brother approaches a stop sign at a velocity of 14 m/s [E] as the light turns amber, He applies the brakes to get the

maximum stopping force while avoiding skidding. The car has a mass of 1500 kg, and the coefficient of friction between the tires and the road is 1.07. Ignoring your brother's reaction time, calculate:
a) The maximum deceleration of the car
b) The stopping distance​
Physics
1 answer:
Viefleur [7K]2 years ago
6 0

The maximum deceleration of the car is 10.5 m/s^{2} approximately. While the stopping distance is 9.3 m

<h3>What is Stopping Force ?</h3>

Stopping force is the force required to stop the object in motion. Stopping force is tantamount to frictional force.

Given that a brother approaches a stop sign at a velocity of 14 m/s [E] as the light turns amber, He applies the brakes to get the maximum stopping force while avoiding skidding. The car has a mass of 1500 kg, and the coefficient of friction between the tires and the road is 1.07. Ignoring your brother's reaction time.

The given parameters are;

  • Velocity V = 14 m/s
  • Mass m = 1500 kg
  • Coefficient of friction μ = 1.07
  • Deceleration a = ?
  • Stopping distance S = ?

The normal reaction N = mg

N = 1500 x 9.8

N = 14700 N

The stopping force = μN = ma

1.07 x 14700 = 1500a

15729 = 1500a

a = 15729/1500

a = 10.486 m/s^{2}

From 3rd equation of motion

V^{2} = U^{2} + 2aS

where U = 0

14^{2} = 2 x 10.486 x S

196 = 20.972S

S = 196/20.972

S = 9.3 m

Therefore, the maximum deceleration of the car is 10.5 m/s^{2} approximately. While the stopping distance is 9.3 m

Learn more about Force here: brainly.com/question/8119756

#SPJ1 ​

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Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
3 years ago
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6 0
3 years ago
A ball at the top of a hill is an example of ______ and a ball rolling down the hill is an example of ______.
larisa [96]

Potential energy , Kinetic energy

<h3 /><h3>What is kinetic energy and potential energy?</h3>

Kinetic energy is a form of energy that posses by an object due to its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.

Potential energy is the energy held by an object due to its position relative to other objects, stresses within itself, its electric charge, or other factors.

According to given condition,

At first,

When a ball is at the top of a hill , it posses a potential energy because the ball is at some height from the surface  where  the energy held by ball is due to its position.

That is why,  a ball at the top of a hill is an example of Potential energy.

Secondly,

When a ball is rolling down the hill , it posses kinetic energy because the ball have some energy that posses by an object due to its motion.

That is why, a ball rolling down the hill is an example of  Kinetic energy.

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6 0
2 years ago
A triangle has two sides of length 10 cm and 14 cm. what can you say about the length of the third side?
Anton [14]

A side of any triangle is less than the sum of the other two sides so the third side must be less than 24cm.

<h3>What is a triangle?</h3>

A triangle is a 3-sided shape that is occasionally referred to as a triangle. There are three sides and three angles in every triangle, some of which may be the same.

The sum of all three angles inside a triangle will be 180° and the area of a triangle is given as (1/2) × base × height.

There are many types of triangles such that right-angle triangles, equilateral triangles, and much more.

In any triangle, the sum of two sides is always greater than the third side.

If the third side is equal to the sum of two sides then it will be a straight line rather than a triangle.

It means 10 + 14 > third side

24 cm > third side

Hence "A side of any triangle is less than the sum of the other two sides so the third side must be less than 24cm".

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3 0
1 year ago
Two very large parallel sheets are 5.00 cm apart. sheet a carries a uniform surface charge density of -9.70 μc/m2 , and sheet b,
MAVERICK [17]
Question is missing. Found on internet the complete text of the problem:

"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."

Solution:

(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span>E= \frac{\sigma}{2 \epsilon _0}
where \sigma is the charge density and \epsilon _0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity.

First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are \sigma_A = -9.70 \mu C/m^2=-9.70\cdot 10^{-6}C/m^2 and \sigma_B = -11.5 \mu C/m^2 = -11.5\cdot 10^{-6} C/m^2.

Sheet a gives an electric field of
E_A= \frac{\sigma_A}{2\epsilon _0}= \frac{-9.7\cdot 10^{-6} C/m^2}{2\cdot 8.85\cdot 10^{-12} F/m} = -5.48\cdot 10^5 V/m
where the negative sign means the field points towards sheet A, in any point of the space.

The electric field produced by sheet B is given by:
E_B = \frac{\sigma_A}{2\epsilon _0}= \frac{-11.5\cdot 10^{-6} C/m^2} {2\cdot 8.85\cdot 10^{-12} F/m} =-6.50\cdot 10^{5} V/m
and again, the negative sign means that the field at any point of the space points towards sheet B.

The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
E=E_1-E_2= -5.48\cdot 10^5 V/m - (-6.50\cdot 10^{5} V/m)=1.02\cdot 10^5 V/m
And the direction is towards sheet B, since it has a field with stronger intensity.

(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
E=E_1+E_2=-11.98\cdot 10^5 V/m
towards right.

(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
E=E_1+E_2=11.98 \cdot 10^5 V/m
towards left.
7 0
4 years ago
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