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DaniilM [7]
3 years ago
7

Question 12 (1 point) Question 12 Unsaved

Physics
1 answer:
Bezzdna [24]3 years ago
7 0

Slope is your answer

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When an electron is displaced in a semiconductor, the hole that's left behind is
morpeh [17]

Answer:

A. attracted to the negative terminal of the voltage source.

Explanation:

When an electron is displaced in a semiconductor, the hole that's left behind is

A. attracted to the negative terminal of the voltage source.

The electron leaving leaves a net + charge, which is attracted to the negative terminal.

8 0
3 years ago
the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0
3 years ago
Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r
Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

v=u-gt

u=gt

t=\dfrac{u}{g}

t=\dfrac{13.7}{9.8}

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

h=ut+\dfrac{1}{2}at^2

Here, a = -g

h=ut-\dfrac{1}{2}gt^2

h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0
4 years ago
From this diagram which of the following can you conclude
Volgvan

Answer: A

Explanation:

I say opposites attract

3 0
3 years ago
Read 2 more answers
flipper (the dolphin) is out in the open ocean hunting tuna avec. he emits his pulse at 22khz and .42 seconds later he hears it
marusya05 [52]
First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
v=\lambda f
where
v is the wave speed
\lambda its wavelength
f its frequency
Using \lambda = 2 cm = 0.02 m and f=22 kHz = 22000 Hz, we get
v=(0.02 m)(22000 Hz)=440 m/s

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
v= \frac{S}{t}= \frac{2L}{t}
and since we know both v and t, we can find the distance L between the dolphin and the tuna:
L= \frac{vt}{2}= \frac{(440 m/s)(0.42 s)}{2}=92.4 m
5 0
4 years ago
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