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tekilochka [14]
2 years ago
11

According to Boyle’s Law, when pressure is increased...

Physics
1 answer:
klasskru [66]2 years ago
4 0

Boyle's law states that when you "shrink"(change its size) a container the pressure increases so the best answer would be D-volume is decreased.

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When light passes through a convex lens, it does this.....
Tpy6a [65]

Explanation:

6. Converge or come together

7. convex

3 0
2 years ago
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Suzie skydiver, who weighs 500 n, reaches terminal velocity of 90 km/h. the air resistance on suzie is then
masha68 [24]
<span>Weight of the skydiver m = 500 N
 Terminal velocity V = 90 km/h
 Here the weight of the person acts as the force, so based on the Newton's third law the applied is the force what we but in the opposite direction making the resistance. So the air resistance exerted on Suzie will be her weight that is 500N</span>
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Within a vacuum, the property to all electromagnetic waves is their
bezimeni [28]
The answer is D) Velocity
7 0
3 years ago
For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee
Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity e of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}

Where:

r_{a} is the apoapsis (the longest distance between the moon and its planet)

r_{p}=0.27 r_{a} is the periapsis (the shortest distance between the moon and its planet)

Then:

e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}

e=\frac{0.73 r_{a}}{1.27 r_{a}}

e=0.57 This is the moon's orbital eccentricity

3 0
3 years ago
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0
2 years ago
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