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Elza [17]
3 years ago
13

Question 7 of 10

Physics
2 answers:
Lyrx [107]3 years ago
8 0

Answer:

A. They can transfer energy through a vacuum

C. They vibrate in two directions that are perpendicular to each other

D. They radiate outward in all directions

Explanation:

UNO [17]3 years ago
6 0

Answer:

A,c,d

Explanation:

Did it on A P E X

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A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, def
Valentin [98]

Answer:

0.2448 point²

Explanation:

1 gry = 1/10 line

1 line = 1/12 inch

=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch

=> 1 inch = 120 gry

1 point = 1/72 inch

=> 1 inch = 72 points

Therefore,

120 gry = 72 points

=> 1 gry = 3/5 point

Therefore,

1 gry² = (3/5)² point²

1 gry² = 9/25 point²

This means that 0.68 gry² will be:

0.68 gry² = 0.68 * 9/25 point²

=> 0.68 gry² = 0.2448 point²

6 0
3 years ago
A 75 kg object is moving west at 5.6 m/s. What is the object's momentum? (p = mv)
Licemer1 [7]

Explanation:

p=mv

p=5.6×75

p= 420

<em>hope</em><em> it</em><em> was</em><em> helpful</em><em> to</em><em> you</em>

7 0
2 years ago
Why is it advised not to hold the thermometer by its bulb while reading it?
Fiesta28 [93]
Because your body heat might change the temperature
7 0
2 years ago
Read 2 more answers
An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in t
Liono4ka [1.6K]

(a) 5.69 N/C, vertically downward

We can calculate the acceleration of the electron by using the SUVAT equation:

d=ut+\frac{1}{2}at^2

where

d = 4.50 m is the distance travelled by the electron

u = 0 is the initial velocity of the electron

t=3.00 \mu s = 3.0 \cdot 10^{-6} s is the time of travelling

a is the acceleration

Solving for a,

a=\frac{2d}{t^2}=\frac{2(4.50)}{(3.0\cdot 10^{-6})^2}=1.0\cdot 10^{12} m/s^2

Given the mass of the electron,

m=9.11\cdot 10^{-31} kg

We can find the electric force acting on the electron:

F=ma=(9.11\cdot 10^{-31})(1.0\cdot 10^{12})=9.11\cdot 10^{-19}N

And the electric force can be written as

F=qE

where

q=-1.6\cdot 10^{-19}C is the charge of the electron

E is the magnitude of the electric field

Solving for E,

E=\frac{F}{q}=\frac{9.11\cdot 10^{-19}}{-1.6\cdot 10^{-19}}=-5.69 N/C

The negative sign means that the direction of the electric field is opposite to the direction of the force (because the charge is negative): since the force has same direction of the acceleration (vertically upward), the electric field must point vertically downward.

(b) Yes

We can answer the question by calculating the magnitude of the gravitational force acting on the electron, to check if it is relevant or not. The gravitational force on the electron is:

F=mg

where

m=9.11\cdot 10^{-31} kg is the mass of the electron

g=9.81 m/s^2 is the acceleration due to gravity

Substituting,

F=(9.11\cdot 10^{-31})(9.81)=8.93\cdot 10^{-30}N

We see that the gravitational force is basically negligible compared to the electric force calculated in part (a), therefore we can say it is justified to ignore the effect of gravity in the problem.

7 0
3 years ago
What kind of data would you display in a bar graph
mr_godi [17]
The data from different subjects, you compare the data too
5 0
3 years ago
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