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3 years ago

A boat was traveling at 15 mph when a passenger threw an object at 10 mph in the same direction the boat was moving. A friend,

Physics

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

the person on the boat is moving 15mph on the boat and throws the ball 10mph. you add that together its 25mph. the other person is standing on land so their is no extra speed.

Explanation:

its common

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2 years ago

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A ball with a weight of 2 N is attached to the end of a cord of length 2m . The ball is whirl in a vertical circle counterclockw

ZanzabumX [31]

Answer:

we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

Explanation:

Weight of the ball is given as

$W = 2N$

so we have

$m = \frac{W}{g}$

$m = 0.204 kg$

now tension force at the top is given as

$T_{top} = 7 N$

$T_{bottom} = 15 N$

Now at the top position by force equation we can say that ball will have two downwards forces

1) Tension force

2) Weight of the ball

so net force on the ball is given as

$F_{net} = T + W$

$F_{net} = 7 + 2 = 9 N$

So we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

8 0

3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

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3 years ago