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3 years ago

Fill in the term that best completes the statement. Work transfers between objects.

Physics

1 answer:

Otrada [13]3 years ago

3 0

Answer:

Energy is the term that best completes the statement.

Hope this helps!

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Neglecting air resistance, what maximum height will be reached by an arrow launched straight upward with an initial speed of 35

tankabanditka [31]

The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)

Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.

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3 years ago

At one instant in time an LR circuit has a current flowing through it of 44mA. At a time 3.3 us later the current is 22mA. The r

irina [24]

B is it because i just took the test

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3 years ago

What is the sentence that we use to remember how to convert metric prefixes?

Marrrta [24]

Answer:

The mnemonic I can use to memorize the metric prefixes in this order is: Gigantic Monsters Killed One Million Men Napping Peacefully. All right, so again, gigantic monsters killed one million men napping peacefully.

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3 years ago

When gravitational forces and air resistance equalize on an object that is falling toward earth and the object stops acceleratin

abruzzese [7]

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3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

2 years ago