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3 years ago

8

A train travelling at 18m/s accelerates uniformly at 2m/s² for 6 seconds. Calculate the final speed of the train.

1 answer:

kobusy [5.1K]3 years ago

8 0

its going to be 30 m/s

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The resistance of moving surfaces produces ________?

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The resistance of moving surfaces produces friction.
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7 0

3 years ago

I need help with this

polet [3.4K]

Exothermic because the heat make part of the result

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3 years ago

A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced s

BARSIC [14]

Answer:

The wavelength is $\lambda_R = 649 *10^{-9}\ m$

Explanation:

From the question we are told that

The wavelength of the red laser is $\lambda_r = 632.8 \ nm = 632.8 *10^{-9}\ m$

The spacing between the fringe is $y_r = 6.00\ mm = 6.00*10^{-3} \ m$

The spacing between the fringe for smaller laser point is $y_R = 6.19 \ mm = 6.19 *10^{-3} \ m$

Generally the spacing between the fringe is mathematically represented as

$y = \frac{D * \lambda }{d}$

Here $D$ is the distance to the screen

and d is the distance of the slit separation

Now for both laser red light light and small laser point D and d are same for this experiment

So

$\frac{y_r}{\lambda_r} = \frac{D}{d}$

=> $\frac{y_r}{\lambda_r} = \frac{y_R}{\lambda_R}$

Where $\lambda_R$ is the wavelength produced by the small laser pointer

So

$\frac{6.0 *10^{-3}}{ 632.8*10^{-9}} = \frac{ 6.15 *10^{-9}}{\lambda_R}$

=> $\lambda_R = 649 *10^{-9}\ m$

7 0

3 years ago

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −

romanna [79]

Answer:

The speed of electron is $1.5\times10^{7}\ m/s$

Explanation:

Given that,

Electric field $E=1.50\times10^{3}\ N/C$

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy $P.E=5.04\times10^{-17}\ J$

Change in potential energy $\Delta P.E=-9.60\times10^{-17}\ J$ as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

$W=-eE\Delta x$

Put the value into the formula

$W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))$

$W=-5.04\times10^{-17}\ J$

We need to calculate the initial velocity

Using change in kinetic energy,

$\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2$

$\Delta K.E=\dfrac{-3mv^2}{8}$

Now, using work energy theorem

$\Delta K.E=W$

$\Delta K.E=\Delta U$

So, $\Delta U=W$

Put the value in the equation

$\dfrac{-3mv^2}{8}=-5.04\times10^{-17}$

$v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}$

Put the value of m

$v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}$

$v=1.21\times10^{7}\ m/s$

We need to calculate the change in potential energy

Using given potential energy

$\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})$

$\Delta U=-4.56\times10^{-17}\ J$

We need to calculate the speed of electron

Using change in energy

$\Delta U=-W=-\Delta K.E$

$\Delta K.E=\Delta U$

$\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}$

Put the value into the formula

$v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}$

$v_{f}=1.5\times10^{7}\ m/s$

Hence, The speed of electron is $1.5\times10^{7}\ m/s$

4 0

3 years ago

If the earth’s tilt on it axis were to increase by 20 degrees, what would happen to earth?

Triss [41]

Each pole would become tropical during summer, but be horribly cold during winter, far colder than they are now. But the equator would be in perpetual twilight during two seasons—and summer the opposing two.

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3 years ago