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3 years ago

Please hurry please

Physics

1 answer:

Ghella [55]3 years ago

4 0

Answer:

0.765m

Explanation:

gravitational potential energy GPE = mass * acceleration due to gravity * height

Given

GPE = 30Joules

Mass m = 4kg

acceleration = 9.8m/s²

Required

Height h

From the formula

h = GPE/mg

h = 30/4(9.8)

h = 30/39.2

h = 0.765m\

Hence the height of the counter is 0.765m

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Answer:

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Explanation:

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When going from a fast speed to a slow speed how is light bent?

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3 years ago

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Here we will calculate the work per unit charge on an electron moving between two potentials. A 9.0 V battery is connected acros

Svet_ta [14]

Answer:

2000 V/m

1778021.69323 m/s

Explanation:

V = Voltage = 9 V = $\Delta V$

q = Charge of proton = $1.6\times 10^{-19}\ C$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

v = Velocity of electron

Electric field is given by

$E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{9}{4.5\times 10^{-3}}\\\Rightarrow E=2000\ V/m$

The electric field is 2000 V/m

Here, the energy of the system is conserved

$\dfrac{1}{2}mv^2=q\Delta V\\\Rightarrow v=\sqrt{\dfrac{2q\Delta V}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 9}{9.11\times 10^{-31}}}\\\Rightarrow v=1778021.69323\ m/s$

The velocity of the electron is 1778021.69323 m/s

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2 years ago

How would taping a coin to the balloon affect the overall motion of the balloon?

Ede4ka [16]

Answer:

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3 years ago

A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

7 0

3 years ago