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2 years ago

A 3.0-kg ball with an initial velocity of (4.0i + 3.0j) m/s collides with a wall and rebounds with a

Physics

1 answer:

irga5000 [103]2 years ago

3 0

The impulse exerted on the ball by the wall is 24i kgm/s.

<h3>Impulse exerted by the ball on the wall</h3>

The impulse exerted by the ball on the wall is the change in the linear momentum of the ball.

J = ΔP

ΔP = Pf - Pi

P = mv

where;

m is mass of the ball
v is the velocity

ΔP = 3(4.0i + 3.0j) - 3(-4.0i + 3.0j)

ΔP = (12i + 9j) - (-12i - 9j)

ΔP = 24i kgm/s

Thus, the impulse exerted on the ball by the wall is 24i kgm/s.

Learn more about impulse here: brainly.com/question/904448

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A circuit has a 9.0 V power supply and two 15 Ω resistors connected in parallel. what is the voltage across each resistor?

Flauer [41]

Answer:

B. 9.0 V

Explanation:

In parallel circuits, the voltage across each circuit is the same across each component, which is also equal to the total voltage of the power supplied. So in this case, the voltage across each resistor is still 9.0V.

The voltage only changes when the resistors can connected in series.

3 0

3 years ago

The property of matter that resists changes in motion is

vredina [299]

Answer:

Inertia

Explanation:

Inertia is the property that any physical object has of remaining in its state of relative motion. Therefore, it is the resistance that opposes matter to modify its state of motion, which includes changes in speed or changes in the direction of movement.

6 0

3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

Which of the following statements are true concerning the creation of magnetic fields? Check all that apply. Check all that appl

Andreas93 [3]

<h3><u>Answer;</u></h3>

A moving electric charge creates a magnetic field at all points in the surrounding region.
An electric current in a conductor creates a magnetic field at all points in the surrounding region.
A permanent magnet creates a magnetic field at all points in the surrounding region.

<h3><u>Explanation;</u></h3>

A magnetic field can be created by running electricity through a wire. All magnetic fields are created by moving charged particles. it is important to also note that charged particles create magnetic fields only when they are moving.
The strength of the magnetic field generated or created is proportional to the amount of current flowing through the wire. Thus, increasing the current increases the strength of the magnetic field.

8 0

3 years ago

In a certain electrolysis experiment involving Al3+ ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many

yawa3891 [41]

Answer:

30.5 x 10³ min.

Explanation:

Atomic weight of aluminium = 27

For the reaction

Al⁺³ + 3e = Al

3 mole of electron is required by 1 mole of aluminium

3 x 96500 C of charge is required for 27 gram of aluminium

60.2 g aluminium will require

$\frac{3\times96500}{27} \times60.2$

645.47 x 10³ C

So charge required = 645.47 x 10³ C

Now Charge = Current x time

Current ( given ) = 0.352A

Time = charge / current = 645.47 x 10³ / .352

=1833 x 10³ s

30.5 x 10³ minutes

6 0

3 years ago