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Law Incorporation [45]

3 years ago

5

A train accelerates from rest and covers a distance of 600m in 20s. What is the train's speed at the end of 20s? What is the fin

al velocity of the train after 20s?

1 answer:

Allushta [10]3 years ago

6 0

Answer:

v = 60 m/s

final velocity after 20 seconds depends on how long it continues to accelerate after 20 seconds. Also we do not know the direction to complete the velocity requirements of a scalar value and direction component.

Explanation:

If we assume that the acceleration is constant

d = ½at² = ½(Δv/t)t² = ½Δvt = ½(v - 0)t = ½vt

v = 2d/t

v = 2(600)/20 = 60m/s

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If X = 5 and Y = 3, what does Z equal? <br> 2X + 2Z = 10Y

Margarita [4]

2(5) + 2 (z) = 10{3}
10 + 2z= 30
2z=20
z=10

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3 years ago

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A ____ is the time required for one half of the nuclei in a radio- ____ isotope to decay.

sukhopar [10]

Answer:

A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.

Explanation:

A radio-active isotope is an isotope which undergoes radioactive decay.

Radioactive decay is a spontaneous process in which the nucleus of an atom changes its state (turning into a different nucleus, or de-exciting), emitting radiation, which can be of three different types: alpha, beta or gamma.

The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.

The law of radio-active decay can be expressed as follows:

$N(t) = N_0 (\frac{1}{2})^{t/t_{1/2}}$

where

N(t) is the number of undecayed nuclei left at time t

N0 is the initial number of nuclei

t is the time

$t_{1/2}$ is the half-life

We see that when $t=t_{1/2}$ (that means, when 1 half-life has passed), the number of undecayed nuclei left is

$N(t) = N_0 (\frac{1}{2})^{t_{1/2}/t_{1/2}}=N_0 (\frac{1}{2})^1=\frac{N_0}{2}$

So, half of the initial nuclei.

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2 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

5 0

2 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

so we can see that distance traveled in each successive second is increasing

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2 years ago

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

Vesna [10]

Answer:

Number of turns per unit length will be $n=2.786\times 10^4turns/m$

Explanation:

We have given that strength of the magnetic field produced by the solenoid B = 7 T

Current in the solenoid i = 200 A

Let the number of turns per unit length is n

Magnetic field due to solenoid is given as $B=\mu ni$ here $\mu$ is permeability of free space n is number of turns per unit length and i is current

So $7=4\pi \times 10^{-7}\times n\times 200$

$n=2.786\times 10^4turns/m$

7 0

3 years ago

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