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Len [333]
3 years ago
14

Stars only emit light within the visible spectrum of the electromagnetic spectrum.

Physics
1 answer:
BabaBlast [244]3 years ago
7 0
A True Stars only emit light within the visible spectrum of the electromagnetic
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the answer do you want me to help you?
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3 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
Two bumper cars colide into each other and each one bounces bach. wich of Newton's laws is it ?
faust18 [17]
Gravity is the answer of newtons law
5 0
2 years ago
Read 2 more answers
A distance of 1.0 meter separates the centers of
professor190 [17]
Neither set of choices is correct.

If the distance is tripled, then the forces decrease to

1/9 Fg. and. 1/9 Fe.

Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.
4 0
3 years ago
What are the (time varying) amplitudes of the E and H fields if summer sunlight has an intensity of 1150 W/m2 in any Town?
Archy [21]

Explanation:

Given that,

Intensity = 1150 W/m²

(a). We need to calculate the magnetic field

Using formula of intensity

I=\dfrac{E^2}{2\mu_{0}c}

E=\sqrt{2\times I\mu_{0}c}

Put the value into the formula

E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}

E=931.17\ N/C

Using formula of magnetic field

B = \dfrac{E}{c}

Put the value into the formula

B=\dfrac{931.17}{3\times10^{8}}

B=0.0000031039\ T

B=3.10\times10^{-6}\ T

(b). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth

We need to calculate the gravitational force

Using gravitational force

F=\dfrac{Gm_{s}M_{e}}{r^2}

Put the value into the formula

F=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}

F=3.522\times10^{22}\ N

We need to calculate the radiation force

Using formula of force

F_{R}=\dfrac{I}{c}\pi\timesR_{E}^{2}

Put the value into the formula

F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.378\times10^{6})^2

F_{R}=4.8\times10^{8}\ N

The gravitational and solar electromagnetic pressure forces of the sun on the earth

\dfrac{F_{G}}{F_{R}}=\dfrac{3.522\times10^{22}}{4.8\times10^{8}}

\dfrac{F_{G}}{F_{R}}=7.3375\times10^{13}

Hence, This is the required solution.

3 0
3 years ago
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