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2 years ago

Someone help me haha;(

Physics

1 answer:

Nikitich [7]2 years ago

6 0

Answer: vf = 51 m/s

d = 112 m

Explanation: Solution attached:

To find vf we use acceleration equation:

a = vf - vi / t

Derive to find vf

vf = at + vi

Substitute the values

vf = 3.5 m/s² ( 8.0 s) + 23 m/s

= 51 m/s

To solve for distance we use

d = (∆v)² / 2a

= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)

= (28 m/s)² / 7 m/s²

= 784 m/s / 7 m/s²

= 112 m

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3 years ago

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Answer:

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Explanation:

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3 years ago

A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ign

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1 year ago

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

2 years ago

Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)

Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

$\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day$

Us the equation of radioactivity

$N=N_{0}e^{-\lambda t}$

$0.1N_{0}=N_{0}e^{-0.08664 t}$

$e^{0.08664 t}=10$

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0

3 years ago