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3 years ago

A piece of metal has attained a velocity of 107.8 m/sec after fallinf for 10 seconds what is its initial velocity

Physics

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

7.8 m/s

Explanation:

Here object is falling with a gravitational acceleration there for we can take acceleration = 10 m/ s² and its constant through out the motion there for we can use motion equation

V = U + at

V - Final velocity

U - Initial velocity

a - acceleration

t - time

V=U+at

107.8=U + 10×10

= 7.8 m/s

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What is the resistance (R) when voltage is 179V and current is 5 Amps?

Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

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2 years ago

The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

8 0

3 years ago

Question #4 Multiple Choice Which of the following is a similarity between an industrial metrologist and a legal metrologist? Bo

sdas [7]

Answer:

choice B

Explanation:i took the test

6 0

2 years ago

Read 2 more answers

g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

5 0

3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago