How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope?.

mina [271]

Answer:

The slope of the graph is what you need. That tells you the speed not the velocity. In order to find the velocity you would also need to know the direction of the motion.

3 0

3 years ago

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I need help with life a child is adopted and I’m a duck where is my left shoe

Pavel [41]

just swim In water and find your shoe

8 0

3 years ago

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A sample of an unknown substance has a mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K

Zinaida [17]

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
345.2 = 89.5(C)(305 - 285)
C = 0.1928 </span>J/g•K

7 0

3 years ago

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someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee

cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) $Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \ in \ the \ journey}$

$Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average speed = 6.00 km/hour

$Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour$

Peter's average speed = 8.48 km/hour

4) $Average \ velocicty = \dfrac{Displacement }{Time \ taken}$

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

$Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Mia's average velocity = 6.00 km/hour

$Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour$

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0

3 years ago

A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular

Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0

3 years ago