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mafiozo [28]
3 years ago
12

To demonstrate an elastic collision, a teacher places a tennis ball on the floor. She wants to hit the ball so that the followin

g conditions are satisfied.
1. After collision, the tennis ball should roll with the velocity of the ball that hit it
2. At the instant of collision, the ball that hit the tennis ball should stop.
What ball should she use?
Golf ball, tennis ball, basketball, or bowling ball?
Physics
1 answer:
ExtremeBDS [4]3 years ago
8 0
Those conditions are met when you have an elastic collision of equal masses. 
So the answer would be a tennis ball. 
Let us write momentum conservation law for this problem. The left-hand side is before the collision and right-hand side is after the collision:
m_1v_1=m_2v_1 
What this equation means is that all momentum of the first ball is transferred to the second ball and velocities before and after the collision are the same.
There is no need to write energy conservation law. We can see that above equation holds true only if m_1=m_2.

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Given that:<br><br> = 2i + 9j ; and ⃗ = -i – 4j . Find . ⃗ ​
Marat540 [252]

Answer:

A.B = -38

Explanation:

A = 2i + 9j and B = -i - 4j.

So, A.B = (2i + 9j).(-i - 4j)

= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)

= -2i.i - 8i.j - 9j.i - 36j.j

since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have

A.B = -2(1) - 8(0) - 9(0) - 36(1)

A.B = -2 - 0 - 0 - 36

A.B = -38

5 0
3 years ago
Solenoid is wound with 2000 turns per meter when the current is 5.2 a, what is the magnetic field within the solenoid?
Drupady [299]

The magnetic field within the solenoid is 0.013T

<h3>What is Solenoid ?</h3>

An electromagnet known as a solenoid creates a regulated magnetic field using a helical coil of wire whose length is significantly higher than its diameter. When an electric current is sent through the coil, it may create a consistent magnetic field inside a defined region of space.

A solenoid operates by creating an electromagnetic field surrounding an armature, which is a moving core. The electromagnetic field causes the armature to move, and when it does, it opens and closes valves or switches, converting electrical energy into mechanical motion and force.

The magnetic field within the solenoid can be calculated by the expression given below.

Write the expression for magnetic field within the solenoid.

B = u₀nI

B = 4π * 10⁻⁷ * 2000 * 5.2

= 0.013 T

to learn more about solenoid go to -

brainly.com/question/1873362

#SPJ4

7 0
1 year ago
A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c
Aleks04 [339]

Answer:

The velocities of the skaters are v_{1} = 3.280\,\frac{m}{s} and v_{2} = 0.024\,\frac{m}{s}, respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b} (1)

Second skater

m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2} (2)

Where:

m_{1} - Mass of the first skater, in kilograms.

m_{2} - Mass of the second skater, in kilograms.

v_{1,o} - Initial velocity of the first skater, in meters per second.

v_{1} - Final velocity of the first skater, in meters per second.

v_{b} - Launch velocity of the meter, in meters per second.

v_{2} - Final velocity of the second skater, in meters per second.

If we know that m_{1} = 70\,kg, m_{b} = 0.043\,kg, v_{b} = 32\,\frac{m}{s}, m_{2} = 58.5\,kg and v_{1,o} = 3.30\,\frac{m}{s}, then the velocities of the two people after the snowball is exchanged is:

By (1):

m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}

m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}

v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}

v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)

v_{1} = 3.280\,\frac{m}{s}

By (2):

m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}

v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}

v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}

v_{2} = 0.024\,\frac{m}{s}

5 0
3 years ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0
3 years ago
The direction that an induced current flows in a circuit is given by
AURORKA [14]

Answer:

Lenz's law

Explanation:

it  states that induced emf  of  different polarities induces a current whose magnetic field opposes the change in magnetic flux through the coil  in order to ensure that original flux is maintained through the coil  when current flows in it.

according to Faraday' s law of electromagnetic induction

Where -ve sign due to lenz's law

Emf is the induced voltage also known as electromotive force

N is the number of loops.

dϕ Change in magnetic flux.

dt Change in time.

8 0
3 years ago
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