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3 years ago

1.Which gas law uses only variables to find a missing variable?

Physics

1 answer:

DanielleElmas [232]3 years ago

5 0

1). D. Combined Gas Law
3.) D. Pressure and temperature
4.) C. Boyles law

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studies show that the ocean surface temperature is increasing which of the following is a possible consequences

katen-ka-za [31]

One possible consequence is that the warmer temperature cause the polar ice to melt even faster

6 0

3 years ago

The magnitude of displacements a and b are 3m and 4m, respectively, c=a+b. What is the magnitude of c if the angel between a and

AysviL [449]

Answer:

(a) 7 m

(b) 1 m

Explanation:

Given:

The magnitude of displacement vector 'a' is 3 m

The magnitude of displacement vector 'b' is 4 m.

The vector 'c' is the vector sum of vectors 'a' and 'b'.

(a)

Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:

$|\overrightarrow c|=|\overrightarrow a|+|\overrightarrow b|\\\\|\overrightarrow c|=3 + 4 = 7\ m$

Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.

(b)

When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

$|\overrightarrow c|=||\overrightarrow a|-|\overrightarrow b||\\\\|\overrightarrow c|=|3 - 4| = 1\ m$

Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.

4 0

2 years ago

A model rocket is launched straight upward with an initial speed of 56.5 m/s. It accelerates with a constant upward acceleration

marta [7]

Answer:

Maximum height reached by the rocket, h = 202.62 meters

Explanation:

It is given that,

Initial speed of the model rocket, u = 56.5 m/s

Constant upward acceleration, $a=1.96\ m/s^2$

Distance traveled by the engine until it stops, d = 198.8 m

Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :

$v=\sqrt{u^2+2ad}$

$v=\sqrt{(56.5)^2+2\times 1.96\times 198.8}$

v = 63.02 m/s

At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.

Again using third equation of motion as :

$v^2-u^2=-2gh$