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Sergeu [11.5K]
3 years ago
9

A 1000.kg car moving at north at 100. km/hr brakes to a stop in 50. m within a uniform acceleration. what are magnitude and dire

ction of the net force on the car ?
thank you so much
Physics
1 answer:
Marianna [84]3 years ago
6 0

Answer:

Force=7.71KN in the opposite direction

Explanation:

Given the mass of the car(M)=1000Kg

The initial speed of the car(u)=100km/hr

we know that 1km/hr=5/18 m/sec

100km/hr=100*5/18 m/sec=27.77m/sec

the distance travelled before it stops (s)=50m

let the acceleration of the car be a

The final velocity of the car is 0.

we know that  v^{2}-u^{2}=2as

0^{2}-27.77^{2}=2(-a)(50)

a=7.712m/sec^{2}

We know that F=Ma

F=1000* 7.712=7712N=7.71KN

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Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A

That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m

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3 years ago
Ten point on important of sports and games in human life​
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3 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

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Answer:

(3) The electrons are attracted to the positively charged nucleus.

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Explanation:

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