COLUMN A

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

4 years ago

A physical property that describes how something feels

Black_prince [1.1K]

Texture hope this helps! :)

5 0

3 years ago

Acceleration usually has the symbol a. It is a vector. What is the correct way

Ghella [55]

Explanation:

a straight line under the letter

3 0

3 years ago

QUICK giving brainlyest

Soloha48 [4]

The momentum of the second ball was 15 kg.m/s.

<h3>What is inelastic collision?</h3>

In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.

In the given question,

Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.

Initial velocity of 1st ball of mass 3 kg is 1 m/s.

So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.

According to Principle of conservation of linear momentum for this inelastic collision,

Initial momentum of first ball + initial momentum of second ball = final momentum of the system

⇒ initial momentum of second ball = final momentum of the system - Initial momentum of first ball

= 18 kg.m/s - 3 kg.m/s.

= 15 kg.m/s.

Hence, initial momentum of second ball = 15 kg.m/s.

Learn more about momentum here:

brainly.com/question/24030570

#SPJ2

5 0

1 year ago

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