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Veseljchak [2.6K]
3 years ago
6

3.0E-17 joules divided by 1.0E -27 kgm/s equals?

Physics
1 answer:
insens350 [35]3 years ago
7 0

Here we have two physical quantities which we have to divide

P_1 = 3.0 \times 10^{-17} J

other is

P_2 = 1.0 \times 10^{-27} kg m/s

now we need to find its division

\frac{P_1}{P_2} = \frac{3.0\times 10^{-17}}{1.0\times 10^{-27}}

now we have

\frac{P_1}{P_2} = 3.0\times 10^{10} Js/kgm

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3 years ago
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When light is directed on a metal surface, the kinetic energies of the photoelectrons a) are random b) vary with the frequency o
jekas [21]

Answer:

b) vary with the frequency of the light

Explanation:

The phone electric effect can be expressed as

K.E=(hv -W•)

Where K.E is the Kinectic energy

W• = work function of the metal

ν =frequency of the radiation

h = Planck's constat

Then, we can see that K.E is proportional linearly to "v" in the equation above.

Therefore, When light is directed on a metal surface, the kinetic energies of the photoelectrons vary with the frequency of the light

5 0
2 years ago
A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg
dezoksy [38]

Answer:

For cast iron we have

h = 0.92 m

For copper

h = 1.05 m

For Lead

h = 1.23 m

For Zinc

h = 2.43 m

Explanation:

As we know that final speed of the block is calculated by work energy theorem

W_f + W_g = \frac{1}{2}mv^2

now we have

-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2

now we have

v^2 = 2gh - 2\mu_k g h cot\theta

v = \sqrt{2gh(1 - \mu_k cot\theta)}

For cast iron we have

4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}

h = 0.92 m

For copper

4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}

h = 1.05 m

For Lead

4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}

h = 1.23 m

For Zinc

4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}

h = 2.43 m

4 0
3 years ago
A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she
klasskru [66]
Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer: 
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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