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3 years ago

15

Velocity is a vector quantity which has both magnitude and direction. Using complete sentences, describe the object's velocity.

Comment on both the magnitude and the direction.

1 answer:

Shtirlitz [24]3 years ago

3 0

Answer & Explanation:

A girl is driving a bicycle at a velocity of 30m/s at 40 degree North west.

Considering the above sentence, the magnitude of the bicycle velocity is 30 meters per second and the direction of the motion of the bicycle is 40 degree North west. So, the bicycle is having both magnitude and direction, so the velocity is a vector quantity.

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An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

lubasha [3.4K]

Answer:

The correct option is

a. v = $g (1-e^{-bt})/b$

Explanation:

Time at which the object start fall t = 0

The acceleration a is given by a = g - bV

Where V = Speed of the object

Speed V² = u² + 2·a·h

However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing

Option a. is the only option that has limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.

v = $g (1-e^{-bt})/b$ as t increases $(1-e^{-bt})$ → 1 s and v→ g/b m/s

6 0

3 years ago

What speed would a fly with the mass of 0.55 g need in order to have the same kinetic energy as the automobile in the term 19

larisa86 [58]

<u>The question does not provide enough information to complete the answer, so I'll assume the needed data to help you to solve your own problem</u>

Answer:

<em>The fly should need to move at 9,534.6 m/s to have the same kinetic energy as the automobile</em>

Explanation:

<u>Kinetic Energy </u>

Is the capacity of a body to do work due to its speed and is computed by

$\displaystyle K=\frac{mv^2}{2}$

We are not given enough data to compare the kinetic energy of the fly with that of the automobile. We'll assume the following characteristics:

$m_a=500\ kg$

$v_a=10\ m/s$

So its kinetic energy is

$\displaystyle K_a=\frac{(500)10^2}{2}$

$\displaystyle K_a=25,000\ J$

The mass of the fly is

$m_f=0.55\ gr=0.00055\ kg$

To have the same kinetic as the automobile:

$\displaystyle \frac{m_fv_f^2}{2}=25,000$

Solving for $v_f$

$\displaystyle v_f=\sqrt{\frac{2(25,000))}{m_f}}$

$\displaystyle v_f=\sqrt{\frac{50,000}{0.00055}}$

$v_f=9,534.6\ m/s$

The fly should need to move at 9,534.6 m/s to have the same kinetic energy as the automobile

5 0

3 years ago

You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from

maxonik [38]

Answer:

A. There is a localization of positive charge near the door handle.

Explanation:

When on a cold morning a person wearing cotton/ polyester cloth walking on the carpet moves toward his car then due to friction between the feet and the carpet there are transfer of electrons from the carpet to our feet, and since our body is a good conductor of electricity the charges spread throughout on the surface of or body.

When the person brings his hands close to the neutral conducting door of the car it gets induced with equal intensity of opposite charge to our hands thus having a concentration of positive charges near to the hand on the car's door is developed as a result of polarization within the conductor.

3 0

3 years ago

How fast does a 2 MeV fission neutron travel through a reactor core?

Artemon [7]

Answer:

The answer is $1.956 \times 10^7\ m/s$

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy $E = mc^2$ , so the definition of energy in this case is

$E = \frac{1}{2}m v^2$ .

From the last equation,

$v= \sqrt{2E/m}$

where

$E = 2 MeV = 3.204 \times 10^{-13} J$

and the mass of the neutron is

$m = 1.675\times 10^{-27}\ Kg$ .

Then

$v = 1.956 \times 10^7\ m/s$

the equivalent of $0.065$ the speed of light.

5 0

3 years ago

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

5 0

3 years ago

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