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Lynna [10]
3 years ago
11

Question 1

Physics
2 answers:
mestny [16]3 years ago
4 0

Answer:

I believe it is C

Explanation:

egoroff_w [7]3 years ago
3 0
C. Transform boundary
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Which of the following best defines speed? (3 points) a It is an object's speed at a specific point in time. b It is the distanc
djverab [1.8K]

Answer:

c.

Explanation:

they are all almost correct.

c is the only fully correct option

6 0
2 years ago
Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the
Leviafan [203]

Answer:

in first case the torque is maximum.

Explanation:

Torque is defined as the product of force and the perpendicular distance.

τ = F x d x Sinθ

In case A: the angle between force vector and the distance vector is 90 so torque is

τ = F x d

In case B: the angle between force vector and the distance is 30°.

τ = F x d x Sin30

τ = 0.5 Fd

So the torque is maximum in first case.

3 0
3 years ago
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

4 0
3 years ago
Number these from least (1) to most (5) mass ?
alisha [4.7K]

Answer:

1. A feather

2. A baseball

3. A small car

4. A truck

5. A large train

Explanation:

5 0
2 years ago
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MIDDLE SCHOOL SCIENCE<br> yee yee please answer 6, 7, and 8
Nikolay [14]

Answer:

what r the questions i can’t see them

Explanation:

5 0
2 years ago
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