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3 years ago

A Velocity vs. time graph is shown What is the acceleration of the object?

Physics

1 answer:

maw [93]3 years ago

7 0

Answer: 4 m/s^2

Explanation: The acceleration of a velocity-time graph is the slope of the line.

The slope of a line is rise over run. In this line, the rise is 4 units and the run is 1 unit. 4/1=4.

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An energy pyramid illustrates that energy in the form of _____ is lost to the surroundings as it's passed from one organism to t

Dima020 [189]

The answer is metabolic heat.

Organisms from the higher trophic levels consume organisms from the lower trophic level and during that process, energy is lost as metabolic heat. Primary producers (plants) contain the greatest amount of energy originally from the sunlight. The next trophic level belongs to primary consumers that consume primary producers. During consumption, energy is lost. Similarly, secondary consumers eat primary consumers and energy is lost again. The highest trophic level is tertiary consumers that contains the least amount of energy.

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3 years ago

Which best describes the way a sound wave is sent through the radio

wlad13 [49]

Answer:

sound wave-electrical wave-radio wave

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3 years ago

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an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

$a_c = \frac{v^2}{r}$

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

$a_c = \frac{(12\ m/s)^2}{2\ m}$

ac = 72 m/s²

Now, the centripetal force applied by the athlete on the hammer will be:

$F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)$

Fc = 504 N

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3 years ago

Students are completing a lab in which they let a lab cart roll down a ramp. The students record the mass of the cart, the heigh

Dennis_Churaev [7]

Answer:

second column

Explanation:

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3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

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3 years ago

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