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3 years ago

Gamma rays have more energy than X-rays. What else is true about gamma rays?

1 answer:

5 0

The answer is D. They have a higher frequency than X-rays.

Gamma-rays falls in the electromagnetic spectrum above, Gamma-rays have frequencies greater than 10¹⁸ cycles per second, or Hertz (Hz)

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Which of the following would be the S I unit to use in measuring the temperature of a hot liquid a. Ampere b. Mole c. Celsius d.

rusak2 [61]

It would be C Celsius just took k12 test

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3 years ago

Read 2 more answers

Which type of wave results from a moderately sloping coastal region?

Naddik [55]

Hey there!

Your answer: Spilling breaker

Spilling breaker usually occurs when a beach or ocean is flat, and as the waves of the wind continues to happen, slowly the region would eventually become a slope.

It's almost like play-dough. Let's say that we set a perfect flat surface of play-dough on the table. As we continue slide our hands one direction, doesn't the play dough have more on one side than the other? It eventually contains a slope if you add enough from the first place.

Your answer: Spilling breaker

8 0

3 years ago

Read 2 more answers

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

What is the difference between the hegelian right, center, and left? islamic?

Marta_Voda [28]

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6 0

3 years ago

A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How

Nata [24]

Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

$W=F.d.Cos \alpha$

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

$F=1000N.2m.Cos(0)=2000N.m=2000 J$

Have a nice day!

8 0

3 years ago