A kite is 100m above the ground. If there are 200m of string out, what is the angle between the string and the horizontal? (Assume that the string is perfectly straight.)

Answers

Answer 1
Answer:

Answer:

the answer is 30°

Explanation:

due to:

sin law of sines

(sin 90)/(200) =(sin\beta )/(100)\narcsin(100(sin90)/(200) )= 30°


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An ocean-going barge is 50.0 m long and 20.0 m wide and has a mass of 145 metric ton. Will the barge clear a reef 1.50 m below the surface of the water?

Answers

Since its mass is 145 metric tons, the barge must displace 145 metric tons of water
in order to float. 145 metric tons= 145,000 kilograms.  Use the rough-but-close rule
of thumb that 1 liter of water has 1 kilogram of mass. So the barge displaces 145,000
liters, which is also 145 cubic meters of water.

(Length) x (width) x (depth of the part under water) = 145 cubic meters

Draft of the barge (depth under water) = 145 / (50 x 20) = 145/70 = 2.07 meters.

No, it will not clear the reef 1.5 meters below the surface.  Either the barge will shave off the top 0.57 meter of the reef, or else the reef will do serious damage to the barge. In any case, there will be a situation.

 

Gravity is affected by

Answers

Gravity is affected by mass and distance

If a 100-N net force acts on a 50-kg car, what will the accelaration of the car be?

Answers

2m/s^2   how to get it: since you already know the force and mass and the formula for force is force=(mass)(acceleration) you isolate the variable of acceleration which is a by dividing (m)(a) by (m) which equals f/m=a then you plug in your known numbers where they go so f=100N m=50kg   100N/50kg= 2 therefore the acceleration is 2m/s^2  hope this helped you!!!

A student pedaling a bicycle applies a net force of 400 N. The mass of the rider and the bicycle is 25 kg. What is the acceleration of the bicycle and the rider?

Answers

The acceleration is 16 m/s^2

Explanation:

We can answer this question by using Newton's second law, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 25 kg is the mass of the rider+bicycle

F = 400 N is the force

Solving for a, we find the acceleration:

a=(F)/(m)=(400)/(25)=16 m/s^2

Learn more about Newton's second law:

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Answer:

The acceleration is

Explanation:

We can answer this question by using Newton's second law, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 25 kg is the mass of the rider+bicycle

F = 400 N is the force

Solving for a, we find the acceleration:

If ∠G=30degrees, g=3, k=4 what is the approximate value of h?2.2 or 4.7

4.7

5.7

Answers

Without a picture of the situation, we don't know the relationship
between angle-G,  g,  k, and 'h'.  We don't even know for sure
whether 'h' is on the same planet as the other things.  No answer
is possible without seeing the picture.

The amplitude of a wave is the height of a wave as measured from the highest point on the wave________ to the lowest point on the wave ________.

Answers

Answer:

Crest to the lowest point on the wave trough

Explanation:

The amplitude of the wave is the height of wave as measured from the highest point on the wave that is known as peak or crest to the lowest point on the wave known as trough. Wave length refers to the length of wave from one peak to another but amplitude or height can be determine by measuring the distance from its crest to its trough.

Final answer:

The amplitude of a wave is the maximum distance, or 'height', from the equilibrium position to the crest (highest point) or trough (lowest point) of the wave. This represents maximum displacement of the wave from its rest position.

Explanation:

The amplitude of a wave refers to the distance from the equilibrium or midpoint of the wave (the rest position) to the highest point of the wave known as the crest, or to the lowest point known as the trough. For instance, considering a wave on a graph representing a sound wave, the crest would be the highest peak on the graph and the trough would be the lowest point. We can visually represent this extent as height, but it's actually a measurement of maximum displacement of the wave from equilibrium (rest position). For physical waves, such as sound or light waves, a larger amplitude leads to a louder sound or a brighter light.

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