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3 years ago

6

A car accelerates from rest at 3.00 m/second squared What is the velocity after 5 secs ? What is the displacement after 5 second

s?

1 answer:

andrey2020 [161]3 years ago

3 0

Explanation:

after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s

The displacement after 5s is

x=vo + (1/2)at²

x = 0 + (1/2)(3m/s²)(5s)(5s)

x= 37.5 m

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Why would it be desirable to use an energy resource that does not produce greenhouse gases ?

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It be desirable to use an energy resource that does not produce greenhouse gases because it contribute to climate change .

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3 years ago

Read 2 more answers

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

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3 years ago

A car traveling east at 92 km/h is passed by a truck moving at 105 km/h. From the point of view of the truck l, how fast is the

vodomira [7]

13km/h all you gotta do is subtract

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3 years ago

A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m

77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}$

$-3i+2j=3.0\times v_{2}$

$v_{2}=-i+0.66j$

The direction of the momentum

$tan\theta=\dfrac{0.66}{-1}$

$\theta=tan^{-1}\dfrac{0.66}{-1}$

$\theta=-33.42^{\circ}$

The direction of the momentum with respect to east

$\theta=180-33.42=146.58^{\circ}$

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

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3 years ago