What would happen if friction suddenly disappears while you are driving a car? (full explanation)

You are riding in a hot air balloon that, relative to the ground, has a velocity of 6.0 m/s due east. You see a hawk moving dire

charle [14.2K]

Answer:Velocity = 6.325m/s

Directional angle= 18.43°

Explanation:

Using Right angle triangle

Let Velocity of ballon&hawk be VHB represent the height of the triangle.

Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.

Let Velocity of hawk and ground BE VHG represent the hypothesis.

Theta = opp/Adj= VHB/VBG

using pythagorean

VHG= SQRT(VHB^2+VBG^2)

VHG= sqrt(2^2+6^2)

VHG= sqrt(4+36)

VHG= 6.325m/s

Tan theta= 2/6

Tan theta =0.3333

Tan^-1 0.3333=18.43°

8 0

3 years ago

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

The velocity after the collision is 2.82 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

6 0

3 years ago

An example of a household appliance with a low and high power rating

olchik [2.2K]

Answer:

Explanation:

There are countless household appliances in every single house. One appliance with a low power rating would be a ceiling fan. On average ceiling fans consume roughly 60w and are found in the majority of houses. On the other side of the spectrum, we have a high power-rating appliance such as a microwave. Microwaves use anywhere between 1000w to 1800w of power in order to function correctly. This is very large power consumption and one of the highest power ratings found for appliances in a household.

5 0

3 years ago

Select all the correct ANSWERS.

MissTica

The ground exerts an equal force on the golf ball.

3 0

4 years ago

An elevator has a mass of 1000 Kg. What force is needed to accelerate it upward at a rate of 2 m/s/s?

zubka84 [21]

The force needed to accelerate an elevator upward at a rate of $2 m / s^{2}$ is 2000 N or 2 kN.

Explanation: 

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as $2 m / s^{2}$ , the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

$\text {Force}=\text {Mass} \times \text {Acceleration}$

$\text { Force }=1000 \times 2=2000 \mathrm{N}=2 \text { kilo Newon }$

So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of $2 m / s^{2}$ .

3 0

3 years ago