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rjkz [21]
3 years ago
15

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

erward? PLEASE HELP
Physics
1 answer:
Ierofanga [76]3 years ago
7 0

2.88 m/s is the velocity afterward.

Explanation:

By using the law of conservation of momentum

Initial momentum = final momentum

\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \times \mathrm{v} \text { equation }(1)

\mathrm{m}_{1}=91.5 \mathrm{kg} \text { is the mass of the first player }\mathrm{m}_{2}=63.5 \mathrm{kg}

\mathrm{m}_{2}=63.5 \mathrm{kg} \text { is the mass of the second player }

\mathrm{u}_{1}=2.73 \mathrm{m} / \mathrm{s} \text { is the initial velocity of the first player (choosing east as positive direction) }

\mathrm{u}_{2}=3.09 \mathrm{m} / \mathrm{s} \text { is the initial velocity of the second player }

v = is their combined velocity afterwards

Solving equation (1) for v

V=\frac{m_{1} u_{2}+m_{2} u_{2}}{m_{1}+m_{2}}

\mathrm{V}=\frac{(91.5 \times 2.73)+(63.5 \times 3.09)}{91.5+63.5}

\mathrm{V}=\frac{(249.7+196.2)}{155}

\mathrm{V}=\frac{445.9}{155}

V = 2.88 m/s

Therefore the velocity afterward is <u>2.88 m/s</u>.

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The speed of sound is 346 m/s. If a sound wave travels at a frequency of 55 Hz, what would its wavelength be?​
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Answer:

6.29 meters.

Explanation:

, where v is the speed of wave and f is the frequency of wave.

We are given that ,

The speed of sound is 346 m/s.

i..e v=346 m/s

A sound wave travels at a frequency of 55 H.

i..e f=55

the wavelength would be 6.29 meters.

This is based on another brainly answer

Link: brainly.com/question/12538018

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A 90-kilogram physics student would weigh 2970 Newtons on the surface of planet X. What is the magnitude of the acceleration due
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<u>Weight = (mass) x (acceleration of gravity)</u>

Divide each side by (mass),and we have

                     Acceleration of gravity = (weight) / (mass)

Acceleration of gravity = 2,970/90 = 33 newtons per kilogram = <em>33 m/s²</em>
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With a mass of 109 kg, Baby Bird is the smallest monoplane ever
OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

  • m_1=61.4kg
  • m_2=109kg
  • v_1=1.84m/s
  • v_2=?

\\ \sf\longmapsto ∆P=P

\\ \sf\longmapsto m_1v_1=m_2v_2

\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}

\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}

\\ \sf\longmapsto v_2=112.976/109

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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
2 years ago
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