Find the value of x. 6x-1 42 10x+5 a 78 x = [?]

A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe

Karolina [17]

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

$F_{r}=kv^2$

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

$F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}$

Put the value into the formula

$3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}$

Put the value of k

$3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}$

$1600-v^2=m\dfrac{d^2v}{dt^2}$

At terminal velocity $\dfrac{d^2v}{dt^2}=0$

So, $1600-v^2=0$

$v=\sqrt{1600}$

$v=40\ ft/sec$

Hence, The velocity is 40 ft/sec.

4 0

3 years ago

Which would likely demagnetize a magnetized iron nail?

olga2289 [7]

The answer to this question is dropping it on a hard surface.

8 0

3 years ago

what would happen when an inflated balloon is place in an ice bath and the atmosphere pressure has not changed?

ruslelena [56]

The inflated balloon shrinks when it is placed in an ice bath with no change in atmospheric pressure.

Explanation:

When the inflated balloon is subjected to an ice bath, it shrinks. This is due to the fact that smaller volume gets occupied by the air/gas inside the balloon as the temperature decreases. Hence, causes the balloon walls to collapse.

An ice bath also lowers the overall air temperature of the balloon inside. As the temperature decreases, the air molecules move more slowly and with lower energy. Because of the particle's lower energy, their collisions with the walls are not enough to keep the inflated balloon.

7 0

3 years ago

Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and

lyudmila [28]

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

K.E₂ = mg(h - 2R)

3 0

3 years ago

Is pulling a dog a contact force or a non contact force

horrorfan [7]

A contact force since you are making physical contact with the dog

Mark my brainliest please

6 0

3 years ago