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3 years ago

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consider the photoelectric effect experiment. in one experiment yellow light shines on a piece of potassium metal. a current is

measured meaning electrons are ejected from the potassium metal. what would happen if the intensity of the yellow light increased

1 answer:

kompoz [17]3 years ago

4 0

Explanation:

If the intensity of the yellow light increased, meaning more photons will strike the Potassium metal per unit area. This will cause more ejection of electrons from the metal and hence, the strength of current will also increase as we know that

I = Q/t, as the charge increase , the current will also increase.

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The speed of light in water is 2.25 x 108 m/s. What is true about the index of refraction of water? A. It is less than 1. B. It

guajiro [1.7K]

Good afternoon!

We can see that $\mathsf{2.25\times10^8\ \textless \ 3\times10^8 = c}$

The index of refraction can be calculated as:

$\mathsf{n = \dfrac{c}{v}}$

As we noted, $\mathsf{v \ \textless \ c}$ , and therefore:

$\mathsf{\dfrac{v}{v} \ \textless \ \dfrac{c}{v}}\\ \\ \\ \mathsf{\dfrac{c}{v} \ \textgreater \ 1}$

Beccause n = c/v, we have:

n > 1

B - Is more than 1.

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3 years ago

Read 2 more answers

Aisha is sitting on frictionless ice and holding two heavy ski boots. Aisha weighs 637 N, and each boot has a mass of 4.50 kg. A

Studentka2010 [4]

Answer:

-0.73 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, in absence of external forces (the ice is frictionless, so no friction), the total momentum of Aisha + two boots is conserved.

At the beginning, their total momentum is zero, since they are at rest:

$p_i = 0$ (1)

After, their total momentum is:

$p_f = Mv + 2mv'$ (2)

where:

M is Aisha's mass

v is Aisha's velocity relative to the ground

m = 4.50 kg is the mass of each boot

v' is the boot's velocity relative to the ground

We can find:

$M=\frac{W}{g}=\frac{637 N}{9.8 N/kg}=65 kg$ is Aisha's mass (where W = 637 N was her weight)

v' can be rewritten as:

$v'=v+6$

because 6 m/s is the velocity of the boots relative to her, while v' is their velocity relative to the ground.

Substituting and combining (1) and (2) we find:

$0=Mv+2m(v+6)\\0=Mv+2mv+12m\\v=\frac{-12m}{M+2m}=\frac{-12(4.50)}{65+2(4.50)}=-0.73 m/s$

and the negative sign indicates that the direction is opposite to that of the boots.

8 0

3 years ago

I can fly but have no wings. I can cry but I have no eyes. Wherever I go, darkness follows me. What am I?

nlexa [21]

cloud?? thats prob wrong lol

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3 years ago

Read 2 more answers

When throwing a ball upwards, the velocity is positive, and the acceleration is negative. True of False?

Setler79 [48]

Answer:

The velocity & acceleration will be taken as negative when a ball is thrown upward because work is done against the gravity.

Explanation:

3 0

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:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

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3 years ago