Upon what does the energy of a quantum depend?

An aeroplane accelerates from rest to 65 m/s for take-off. It travels for

Virty [35]

Answer:

1.76m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity = 65m/s

Distance traveled = 1200m

Unknown:

Acceleration = ?

Solution:

This is linear velocity and we apply the appropriate motion equation to solve this problem.

V² = U² + 2as

S is the distance

u is the initial velocity

V is the final velocity

a is the acceleration

Now, insert the parameters and solve;

65² = 0² + 2 x a x 1200

4225 = 2400a

a = 1.76m/s²

5 0

2 years ago

Which of the following have derived units? A. 56 kg B. 2.5 m C. 87 m/s² D. 60 N

photoshop1234 [79]

Answer:

The answer is C!!!!!!!

Becuz meters and seconds are derived into m/s²

Explanation:

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Thanks :-)

6 0

2 years ago

Read 2 more answers

What is a supernova?

kozerog [31]

A supernova is a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.

8 0

2 years ago

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

3 0

3 years ago