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3 years ago

The velocity of a 10.0 kg object that has 720 J of kinetic energy is m/s. (Report the answer to two significant figures.)

Physics

2 answers:

RideAnS [48]3 years ago

7 0

the answer is 12 m/s

nikdorinn [45]3 years ago

4 0

Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. It is expressed as:

KE = mv^2 /2

720 = 10.0v^2 /2
v = 12 m/s

Hope this answers the question. Have a nice day.

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andrey2020 [161]

Answer:

400 milli-rems

Solution:

As per the question;

Maximum energy of particle, $E_{m} = 0.0186\ MeV$

Weight, w = 1 kg

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Now,

Equivalent dose is given by:

$D_{eq} = \frac{E}{w} =\frac{4\times 10^{- 3}}{1} = 4\times 10^{- 3}\ J/kg$

1 Gy = 1 J/kg

Also,

1 Gy = $1\times 10^{5}\ milli-rems$

Therefore,

Dose equivalent in milli-rems is given by:

$D_{eq} = 4\times 10^{- 3}\times 1\times 10^{5} = 400\ milli-rems$

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3 years ago

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Elena L [17]

Answer:

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3 years ago

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Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

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3 years ago

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

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4 years ago

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Brrunno [24]

Answer:

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