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nalin [4]
3 years ago
9

All diseases can be prevented. A) True or B) False

Physics
2 answers:
lakkis [162]3 years ago
8 0
B. false is the answer.
ad-work [718]3 years ago
6 0
The correct answer is B)False
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A plane crashes with a deceleration of 185 m/s. How many g’s is this?
Afina-wow [57]

Answer:

26 g's

Explanation:

Hope this helps~

Have a great day

Zero

3 0
3 years ago
Comparing the speed of sound in liquids, gases, and solids, the speed of sound is usually lowest in ( ) and highest in ( ).
nydimaria [60]

Answer:

the answer is the 4th one. the speed of sound is lowest in gas and highest in solid and its because the molecules in solid is closer to each other than the molecules in gas and when the molecules are close to each other, they allow the waves of the sound to travel more quickly.

7 0
3 years ago
What is 3.75 x 10^-7?
pogonyaev

Answer:

Explanation:

3.75 * 10^-7

=3.75 * 1/10^7

=3.75/10000000

=3/800000000

any base which has it's power negative do it's reciprocal then the power will be positive.

8 0
3 years ago
Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24
Oksi-84 [34.3K]
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}
Now At xp we have:
\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0
\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0
Which is a second order equation, using the quadratic formula to solve for xp would give us:
xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
or
xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0
4 years ago
The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is
yKpoI14uk [10]

We use the voltage division problem  between the load resistance, amplifier output resistance as

V_{out} = V_{amlifire} \times \frac{R_{load} }{R_{load} + R_{out} }.

Here, V_{out} is the output voltage, V_{amlifire} is the amplifier voltage, R_{load} is the load resistance and R_{out} is the amplifier output resistance.

Therefore,

1-\frac{20}{100} = \frac{1 \ k\Omega }{1 \ k\Omega +R_{out} } \\\\ R_{out} = \frac{1 \ k\Omega }{0.8} -1 \ k\Omega =1250 \Omega -1000 \Omega =250 \Omega.

Thus, the amplifier output resistance is 250 \ \Omega.

4 0
3 years ago
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