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1 year ago

A 85-kg physics student jumps from a dock into a 135 kg boat that is at rest by the dock. If the velocity

Physics

1 answer:

Gnesinka [82]1 year ago

4 0

Answer:

The final velocity of the student and the boat is 1.66 m/s

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is P=mv. If we have a system of bodies, then the total momentum is the sum of them all

$P=m_1v_1+m_2v_2+...+m_nv_n$

Let's call $m_s = 85\ kg$ the mass of the physics student, $m_b = 135\ kg$ the mass of the boat, $v_{so} =4.3\ m/s$ the initial speed of the student, $v_{bo} =0$ the initial speed of the boat, $v_f$ the final speed of both, assumed common since they keep joined after the jump. Applying the law of conservation of momentum, being $P_o\ and\ P_f$ the initial and final momentum of the system, we have

$P_o=P_f$

$m_sv_{so}+m_bv_{bo}=(m_s+m_b)v_f$

Solving for $v_f$

$v_f=\frac{m_sv_{so}+m_bv_{bo}}{(m_s+m_b)}$

$v_f=\frac{85(4.3)+0}{(85+135)}$

$v_f=1.66 \ m/s$

The final velocity of the student and the boat is 1.66 m/s

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A spherical conductor has a radius of 14.0 cm and a charge of 42.0 µC. Calculate the electric field and the electric potential a

Yakvenalex [24]

Answer:

0 MN/C, 2.697 MV

4.1953 MN/C, 1.2586 MV

19.2642857143 MN/C, 2.697 MV

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field at r = 8 cm

E = 0 (inside)

Electric potential is given by

$V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV$

Electric potential is 2.697 MV

r = 30 cm

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C$

Electric field is 4.1953 MN/C

$V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV$

Electric potential is 1.2586 MV

r = R = 14 cm

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C$

The electric field is 19.2642857143 MN/C

$V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV$

The potential is 2.697 MV

7 0

1 year ago

50 POINTS!!! PLEASE ANSWER CORRECTLY!

Talja [164]

Guglielmo Marconi invented the first radio transmitter.

Since he is not one of the options, I would go with B. Hertz.

Marconi started his research after Hertz’s experiments and work in electromagnetic waves.

4 0

1 year ago

a student has revived a D on her english paper and is discussing with her teacher strategies that she might use to achieve highe

noname [10]

Interpersonal communication.

It's a conversation between two people and no one else.

8 0

1 year ago

As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air

nignag [31]

Answer:

B.-80.35 J

i dont know the calculation

8 0

1 year ago

a telescope is oriting on a spacecraft aroudn the earth. Speed of 3.25 x 10^5 mass 7500 kg. What is the de Broglie wavelength of

Scrat [10]

Answer:

2.72 x 10^-43 m

Explanation:

mass of the telescope = 7500 kg

speed of the telescope = 3.25 x 10^5 m/s

de Broglie's wavelength of the telescope is given as

λ = h/mv

where

λ is the wavelength of the telescope

h is the plank's constant = 6.63 × 10-34 m^2 kg/s

m is the mass of the telescope = 7500 kg

v is speed of the telescope = 3.25 x 10^5 m/s

substituting value, we have

λ = (6.63 × 10-34)/(7500 x 3.25 x 10^5)

λ = 2.72 x 10^-43 m

8 0

1 year ago