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4 years ago

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A lead pellet of mass 10.0g is shot horizontally into a stationary wooden block of mass 100g. The pellet hits the block with an

impact velocity of 250ms^–1. It embeds itself in the block and it does not emerge.
What will be the speed of the block immediately after the pellet is embedded?

1 answer:

Roman55 [17]4 years ago

6 0

Hope this helps you.

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Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

6 0

4 years ago

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity (in m/s) must a basket ball player l

mamaluj [8]

Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

$v^2-u^2=2ah$

here Final Velocity v=0

a=acceleration due to gravity

$0-u^2=2\left ( -g\right )h$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.81\times 1.24}$

$u=\sqrt{24.328}$

u=4.93 m/s

7 0

4 years ago

An infrared (IR) and ultraviolet (UV) wave propagating through a vacuum must have the same____

Zinaida [17]

Answer:

D. Wavelength

Explanation:

An infrared (IR) and ultraviolet (UV) wave propagating through a vacuum must have the same wavelength.

3 0

3 years ago

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity = 1.35 mm

λ = wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2

=5.0625 *10^-7 = 506.25

so

Wavelength of the light 506.25.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

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1 year ago

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

3 years ago