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3 years ago

This graph shows velocity vs. time. What does the slope of the line represent? A. speed B. force C. acceleration D. distance

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

I believe the answer is C. Acceleration. (:

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A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t

emmainna [20.7K]

Efficiency = Work Output / Work Input

92% = Work Output / 100

0.92 = Work Output / 100

Work Output = 0.92 * 100

Work Output = 92 joules.

8 0

3 years ago

If Phoenix, Arizona, experiences a cool, wet day in June, does that mean the regions climate is changing?

jenyasd209 [6]

It doesn't mean that the climate is changing, it is probably just the morning dew.

5 0

3 years ago

A student standing on a knoll throws a snowball horizontally 4.5 meters above the level ground toward a smokestack 15 meters awa

Elan Coil [88]

Answer:

2.4 m

Explanation:

Consider the motion along the vertical direction

$y_{o}$ = initial position of ball above the ground = 4.5 m

$t$ = time taken by the ball to hit the smokestack = 0.65 s

$v_{oy}$ = initial velocity of the ball along vertical direction

$a_{y}$ = acceleration due to gravity = - 9.8 m/s²

$y$ = position of ball at the time of hitting the smokestack

Using the kinematics equation

$y = y_{o} + v_{oy} t + (0.5) a_{y} t^{2}$

inserting the above values

$y = 4.5 + (0) (0.65) + (0.5) (- 9.8) (0.65)^{2} \\y = 2.4 m$

6 0

3 years ago

CNA Secondary and Preparatory School

ale4655 [162]

Answer:

2:13 2) cm' c6462.

Explanation:

3 0

2 years ago

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field is 1.71436 m

4 0

3 years ago

Read 2 more answers