A wave travels with speed 200 m/s. Its wave number is 1.5 rad/m. What are its (a) wavelength and (b) frequency

Every second, the Sun converts about 600 million tons of hydrogen into 596 million tons of helium. The remaining 4 million tons

Whitepunk [10]

Answer:

Converted to an amount of energy equal to 4 million tons times the speed of light squared. ejected into space in a solar wind.

Explanation:

The 4 million tons of mass is converted to the amount of energy that is equal to 4 million tons times the speed of light squared. This energy moves from the sun with the help of solar winds and received by the planets present in the solar system. This solar energy moves in the form of solar radiation because there is no medium for propagation so that's why we can say that the mass is converted into energy that moves in the form of radiation in discrete packets.

4 0

2 years ago

D section a only still need help on this

SIZIF [17.4K]

Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.

The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.

Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.

5 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

When atoms are split, they release energy. this concept applies to

balu736 [363]

This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.

6 0

3 years ago

An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu

romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

qnet = (Iin - Iout)*t

qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

V = k*qnet / r

Where, k = 8.99*10^9 ..... Coulomb's constant

qnet = V*r / k

t = 1000*0.1 / (8.99*10^9 * 0.000002)

t = 5.56 ms

7 0

3 years ago