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3 years ago

A 1000 kg car makes a turn on a banked curve. The radius is 300 m.

Physics

1 answer:

Alenkinab [10]3 years ago

5 0

b) The maximum speed of the car is 47.0 m/s

a) The centripetal force on the car is 7363 N

Explanation:

We have to proceed by solving part b) first. The frictional force between the tires of the car and the road provides the centripetal force that keeps the car in circular motion, therefore we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

$\mu = 0.75$ is the coefficient of friction

m = 1000 kg is the mass of the car

r = 300 m is the radius of the turn

v is the speed of the car

The maximum speed of the car is the speed for which the frictional force is still enough to keep the car in circular motion, so

$\mu mg \geq m\frac{v^2}{r}$

Therefore, solving for $v$ , we find

$v = \sqrt{\mu g r}=\sqrt{(0.75)(9.8)(300)}=47.0 m/s$

The centripetal force for an object in circular motion is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the object

v is its speed

r is the radius of the circular path

For the car in this problem,

m = 1000 kg

v = 47.0 m/s

r = 300 m

Solving for F, we find the force:

$F=(1000) \frac{47.0^2}{300}=7363 N$

Learn more about circular motion:

brainly.com/question/2562955

#LearnwithBrainly

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A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge

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Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is $\lambda$ and the net charge per unit length is $2 \lambda$ .

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let $\lambda_i\ and\ \lambda_o$ are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

$\phi=\dfrac{q_{enclosed}}{\epsilon_o}$

$E.A=\dfrac{q_{enclosed}}{\epsilon_o}$

$0=\dfrac{\lambda_i+\lambda}{\epsilon_o}$

$\lambda_i=-\lambda$

For outer surface,

$\lambda_i+\lambda_o=2\lambda$

$-\lambda+\lambda_o=2\lambda$

$\lambda_o=3\lambda$

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

$E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}$

$E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}$

Hence, this is the required solution.

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2 years ago

A ceiling fan has three blades. The moment of inertia of a blade is 0.2kgm^2. The net torque exerted on fan blades is 8Nm. Find

olchik [2.2K]

Answer:

(A) the angular acceleration of the blades is 13.33 m/s.

Explanation:

Given;

moment of inertia of a blade, I = 0.2 kgm²

net torque exerted on fan blades, ∑τ = 8Nm

Torque is given as product of moment of inertia and angular acceleration;

τ = Iα

where;

α is the angular acceleration

Since there are three blades of the ceiling fan, the net torque is given as;

∑τ = (3I)α

∑τ = 3Iα

α = ∑τ / 3I

α = (8) / (3 x 0.2)

α = 13.33 m/s

Therefore, the angular acceleration of the blades is 13.33 m/s.

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2 years ago

At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar

swat32

Answer:

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

Explanation:

First of all let's define the specific molar heat capacity.

$C = \frac{-Q}{n\cdot \Delta T}$ (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system

Now, we can find n with the molar mass (M) the mass of the compound (m).

$n=\frac{m}{M}=6.95\cdot 10^{-3} moles$

Using (1) we have:

$C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}$

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

I hope it helps!

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3 years ago

Billiard balls moving across a pool table is an example of _____ friction.

frozen [14]

It is an example of rolling friction because balls roll.

Answer is ROLLING

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2 years ago

A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having

Mrac [35]

Answer:

F_Balance = 46.6 N ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

∑ F = 0

Fe –W + F_Balance = 0

F_Balance = - Fe + W

The electric force is given by Coulomb's law

Fe = k q₁ q₂ / r₂

The weight is

W = mg

Let's replace

F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

q₁ = + 8 μC = +8 10⁻⁶ C

q₂ = - 3 μC = - 3 10⁻⁶ C

r = 0.3 m = 0.3 m

Let's calculate

F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²

F_Balance = 49 - 2,397

F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

m' = F_Balance / g

m' = 46.6 /9.8

m' = 4,755 kg

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3 years ago