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bogdanovich [222]
4 years ago
13

Which would cause the greatest increase in the acceleration of a satellite?

Physics
2 answers:
vagabundo [1.1K]4 years ago
6 0
Its C. a decrease in the radius and an increase in the tangential speed ,I'm 100% sure
nekit [7.7K]4 years ago
5 0

Answer:

A decrease in the radius and an increase in the tangential speed

Explanation:

As we know that in circular motion the acceleration towards the center of the circular path is given by the formula

a_c = \frac{v^2}{R}

now we can say that in order to increase the acceleration of the circular motion we can change the value of speed v and radius R both

here we know that

v = tangential speed

R = radius of path

so here since speed is directly dependent on the acceleration while radius is inversely depends on the acceleration so correct answer must be

A decrease in the radius and an increase in the tangential speed

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Juan is making ice tea. When he adds ice to the tea, why does the tea cool down?​
Tresset [83]

Answer: because the ice was cold, and it melted in the tea. So the tea is now cooler.

Explanation:

4 0
3 years ago
Can someone help me?!!!!!
german
<h2>Hello!</h2>

The answer is:

The first option,  the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

Distance=NorthCoveredDistance+EastCoveredDistance

Distance=4*180m+3*180m=720m+540m=1260m

Actual distance between the start and the end point (displacement):

ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0
3 years ago
PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat
Svetach [21]
Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of  1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C
4 0
4 years ago
How r u?<br> Are u good?
Maslowich

Answer:

i'm good, hru hope ur staying safe

Explanation:

6 0
3 years ago
Read 2 more answers
A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

  • u is the initial velocity = 97 km/h = 26.94 m/s

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

μ = (7.72)/9.8

μ = 0.78

Learn more about coefficient of friction here: brainly.com/question/14121363

#SPJ1

5 0
2 years ago
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