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Greeley [361]
4 years ago
10

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

ocity of +4.2 m/s, grabs and holds onto him so that they move off together with a velocity of +2.3 m/s. If the mass of the tackler is 125 kg, determine the mass of the receiver in kilograms. Assume momentum is conserved.
Physics
1 answer:
Fudgin [204]4 years ago
4 0

Answer:103.261kg

Explanation:

Assume m to be mass of the receiver

Momentum before =(125*4.2)-m*0

Momentum after =2.3(125+m)

525=287.5+2.3m

m=237.5/3.3

m=103.261kg

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A wave has a wavelength of 14 m and a frequency of 24 Hz, what is the speed of the wave?
lapo4ka [179]

Answer:

Explanation: Speed = Wavelength x Wave Frequency. In this equation, wavelength is measured in meters and frequency is measured in hertz (Hz), .

6 0
3 years ago
A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −
romanna [79]

Answer:

The speed of electron is 1.5\times10^{7}\ m/s

Explanation:

Given that,

Electric field E=1.50\times10^{3}\ N/C

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy P.E=5.04\times10^{-17}\ J

Change in potential energy \Delta P.E=-9.60\times10^{-17}\ J as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

W=-eE\Delta x

Put the value into the formula

W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))

W=-5.04\times10^{-17}\ J

We need to calculate the initial velocity

Using change in kinetic energy,

\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2

\Delta K.E=\dfrac{-3mv^2}{8}

Now, using work energy theorem

\Delta K.E=W

\Delta K.E=\Delta U

So, \Delta U=W

Put the value in the equation

\dfrac{-3mv^2}{8}=-5.04\times10^{-17}

v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}

Put the value of m

v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}

v=1.21\times10^{7}\ m/s

We need to calculate the change in potential energy

Using given potential energy

\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})

\Delta U=-4.56\times10^{-17}\ J

We need to calculate the speed of electron

Using change in energy

\Delta U=-W=-\Delta K.E

\Delta K.E=\Delta U

\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}

Put the value into the formula

v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}

v_{f}=1.5\times10^{7}\ m/s

Hence, The speed of electron is 1.5\times10^{7}\ m/s

4 0
3 years ago
Solve this question with steps
Novay_Z [31]
The problem was too big to type in my phone.

I hope my answer is readable.

P.S After the collision P is also moving in the same direction as Q.

5 0
4 years ago
Think about the types of forces holding the atoms together in different chemicals. Are there any patterns that can be drawn from
Leya [2.2K]

Hello. You did not enter the data to which this question refers, which makes it impossible for it to have an exact answer. However, I will try to help you in the best possible way.

The forces that hold the elements together are called intermolecular forces. They are formed by covalent bonds between the molecules and can be called: dipole-induced (occurs between nonpolar molecules that have a negative pole and a positive pole) and dipole-dipole (occurs between polar moileculas, except when hydrogen is present).

5 0
3 years ago
A student went to a hill station early in the morning, he could hear the echo of his clap after 0.1 second. When he went to the
Dahasolnce [82]

Answer:

See the explanation.

Explanation:

The speeds of sound depend upon the temperature of the medium. As the temperature increased in the afternoon, the speed of sound increases. Then the time taken by reflected sound will be less than 0.1 sec in the afternoon. And to hear an echo the time gap between an original sound and reflected sound must be at least 0.1 sec. That is why the student could not hear the echo at all in the afternoon.

7 0
3 years ago
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