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3 years ago

What is the velocity of a bicycle in meters per second if it travels 1 kilometer west in 4.1 minutes

1 answer:

6 0

1 kilometre is equal to 1000m
and 4.1 minutes is equal to 246 seconds
thus 1000/246 = 4.065 m/s
and the direction is towards the west

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Can viruses, bacteria, and fungi be considered parasites?

Pani-rosa [81]

Yes, they live off of other organisms and harm the organisms.

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3 years ago

Label these as balanced or unbalanced:

kvasek [131]

Answer:

A book lying on a table - Balanced force
An airplane cruising in level flight - Balanced
A rock falling from a cliff - Unbalanced force
A bridge collapsing in an earthquake - Unbalanced force
A man sitting on a park bench - Balanced force
A space shuttle taking off - Unbalanced force
A car maintaining a constant speed on a straight road - Balanced force
An airplane landing - Unbalanced force

Explanation:

Usually, one or more forces act on a body at an instant of time. When these forces acting on a body and bring the body in the equilibrium position, the force is said to be balanced. The unbalanced force changes the equilibrium state of the body.

As in the case of an airplane cruising in a level flight, the weight of the plane will be equal to the lift force and the thrust is equal to the drag. So the plane is experiencing a balanced force.

4 0

3 years ago

Read 2 more answers

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2× $10^{-6}$ J

(b) KE = 2× $10^{-6}$ J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2× $10^{3}$ N/C

q = 2× $10^{-9}$ C

d = 50 - (-30) = 80× $10^{-2}$ = 8× $10^{-1}$ m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2× $10^{3}$ . 2× $10^{-9}$ . 8× $10^{-1}$

ΔU = 3.2× $10^{-6}$ J

(b) Mechanical Energy is constant, so:

$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2× $10^{3}$ . 2× $10^{-9}$ . 5× $10^{-1}$ )

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0

3 years ago

USE K U E S

nata0808 [166]

Explanation:

Problem 1.

Initial speed of the runner, u = 0

Acceleration of the runner, $a=4.2\ m/s^2$

Time taken, t = 100 s

Let v is the speed of the runner now. Using the first equation of kinematics as :

$v=u+at$

$v=at$

$v=4.2\ m/s^2\times 100\ s$

v = 420 m/s

Problem 2.

Initial speed of the plane, u = 0

Distance covered, d = 300 m

Time taken, t = 25 s

Using the equation of kinematics as :

$d=ut+\dfrac{1}{2}at^2$

$d=\dfrac{1}{2}at^2$

$a=\dfrac{2d}{t^2}$

$a=\dfrac{2\times 300\ m}{(25\ s)^2}$

$a=0.96\ m/s^2$

Problem 3.

A ball free falls from the top of the roof for 5 seconds. Let it will fall at a distance of d. It is given by :

$d=ut+\dfrac{1}{2}gt^2$

$d=\dfrac{1}{2}\times 9.8\times (5)^2$

d = 122.5 meters

Let v is the final speed at the end of 5 seconds. Again using first equation of kinematics as :

$v=u+gt$

$v=9.8\ m/s^2\times 5\ s$

v = 49 m/s

Hence, this is the required solution.

3 0

3 years ago

How could you use the barometer to determine the height of a building?

Nady [450]

Take the barometer to the roof of the building. Drop the barometer from the edge of the roof, and, with your wrist watch or a stop watch, measure the time it takes for the barometer to hit the ground or the street below. Then the height of the roof, in meters, is 4.9 times the square of the time in seconds.

5 0

3 years ago