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3 years ago

True or false?

Physics

2 answers:

Ann [662]3 years ago

8 0

Answer: true

Explanation:

cricket20 [7]3 years ago

6 0

Answer:

True is the correct answer.

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An astronaut rides in a circular training accelerator at 16 m/s. If the radius of the accelerator is 8.0 meters, what is his cen

iren [92.7K]

Answer:

$a=32\ m/s^2$

Explanation:

Given that,

The velocity of an astronaut in a circular path, v = 16 m/s

The radius of the accelerator, r = 8 m

We need to find his centripetal acceleration. The formula that is used to find the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(16)^2}{8}\\\\a=32\ m/s^2$

So, the required centripetal acceleration is $32\ m/s^2$ .

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3 years ago

A particle is moving along the x-axis so that its position at any time t is greater than and equal to 0 is given by x(t)=2te^-t?

erastovalidia [21]

For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left

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3 years ago

What is frictional less pulley

Talja [164]

Answer:

Conditions under which the belt and pulleys are operating – The friction between the belt and pulley may decrease substantially if the belt happens to be muddy or wet, as it may act as a lubricant between the surfaces.

Explanation:

I hope that this would be helpful

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3 years ago

What is meant by an acceleration of negative 2metre per second square

soldi70 [24.7K]

means that a body is in motion, and its velocity is measured in meters per second. And, that velocity is increasing by two meters per second, every second.

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3 years ago

Read 2 more answers

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

8 0

3 years ago