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2 years ago

6

Describe how a wheel and axle work?

1 answer:

andrezito [222]2 years ago

7 0

A weally works when you pull something and a axle is something that you can push

Explanation:

I hope this helps

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A wave is moving at 18 m/s. If its wavelength is 3 meters, what is its frequency?

Vlada [557]

here's the solution,

we know that,

=》

$wave \: speed = wavelength \times frequency$

so,

=》

$18 = 3 \times f$

=》

$f = \dfrac{18}{3}$

=》

$f = 6$

frequency = 6 hertz

7 0

3 years ago

Yuliya22 [10]

PE= 3kg x 10N/kg x 10m
= 300J

8 0

3 years ago

How do we know that pulsars must be neutron stars?

slavikrds [6]

Answer:

funny

Explanation:

3 0

3 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

3 years ago

Read 2 more answers

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

3 years ago