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Gwar [14]
2 years ago
13

PLEASE HELP!!! THANKS I GIVE BRAINLIEST !!A student examines the effect of the number of D batteries in a closed circuit on the

brightness of a light bulb. In the experiment, four circuits were built with 1, 2, 3, and 4 batteries respectively. For each trial, the brightness of the light bulb was measured using a light meter. Which variable should the student keep constant
Physics
1 answer:
kirza4 [7]2 years ago
4 0

Answer:

The batteries

Explanation:

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A cloud is drifting across the sky at a constant velocity of 724metersperminute to the east. It is at an altitude of 9,200 meter
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Answer:

The cloud moves 9050 meters to the east in 12.5 minutes.

Explanation:

Let suppose that mass of the cloud is negligible. meaning that effects of gravity are negligible and that altitude of the cloud remains constant. If the cloud drifts at constant velocity, travelled distance is defined by following formula:

\Delta s = v\cdot \Delta t (1)

Where:

v - Velocity, in meters per second.

\Delta t - Time, in seconds.

If we know that v = 724\,\frac{m}{min} and \Delta t = 12.5\,min, then the travelled distance after 12.5 minutes is:

\Delta s = 9050\,m

The cloud moves 9050 meters to the east in 12.5 minutes.

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What does a pshycometer measure
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WHAT IS THE POTENIAL ENERGY OF A 3 KILOGRAM BALL THAT IS ON THE GROUND
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Answer:

147.15 Joules.

Explanation:

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You throw a ball vertically upward so that it leaves the ground with velocity +5.00 m/s. (a) What is its velocity when it reache
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Answer:

(a) The velocity (v)  of the ball when reaches its maximum altitude is zero .

     v= 0

(b) The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball  is  g =-9,8 m/s².

(c)velocity of the ball when it returns to the ground :

v_{f} =5 \frac{m}{s} in direction -y

v_{f} =-5 \frac{m}{s}

(d)a=g = -9,8 m/s² : acceleration of the ball when it returns to the ground

Explanation:

Ball Kinematics

We apply the free fall formula:

vf²=v₀²+2*a*y Formula (1)

y:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration

Data

v₀ = +5.00 m/s

Problem development

(a) What is its velocity when it reaches its maximum altitude?

in ymax,  vf=0 , At maximum height the velocity is zero and the ball falls freely

(b) What is its acceleration at this point?

The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball  is 9,8 m/s².

g= -9.8 m/s²

c) What is the velocity with which it returns to ground level?

We apply the Formula (1) to calculate the maximum height (h):

Ball movement up

vf²=v₀²+2*a*h

0 = (5)² +2*(-9.8)*h

19.6*h = 25

h= 25/19.6 = 1.275 m

Ball movement down

The distance the ball goes up is equal to the distance it goes down

h= 1.275 m , v₀=0

vf²=v₀²+2*a*h

vf²=0+2*(-9.8)*(-1.275)  

v_{f} = \sqrt{25}

v_{f} =5 \frac{m}{s} in direction -y

v_{f} =-5 \frac{m}{s}

(d) What is its acceleration at this point?

a is constant , a= g = -9.8 m/s²

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