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3 years ago

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An outfielder with a mass of 86 Kilograms catches a baseball with a mass of 0.15 kilograms.The velocity of the ball just before

being caught is 40 meters per second.What is the impulse imparted by the ball onto the outfielder?
A. 6.00 Ns
B. 3,440 Ns
C. 2.00 Ns
D. 3,434 Ns

1 answer:

nordsb [41]3 years ago

5 0

Answer:

C

Explanation:

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An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init

elena-14-01-66 [18.8K]

Answer:

Approximately $71\%$ .

Explanation:

The formula for the kinetic energy $\rm KE$ of an object is:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ ,

where

$m$ is the mass of that object, and
$v$ is the speed of that object.

Important: Joule ( $\rm J$ ) is the standard unit for energy. The formula for $\rm KE$ requires two inputs: mass and speed. The standard unit of mass is $\rm kg$ while the standard unit for speed is $\rm m \cdot s^{-1}$ . If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

$m = 63\; \rm g = 0.063\; \rm kg$ .

Initial $\rm KE$ of this arrow:

$\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}$ .

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

$\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%$ .

8 0

3 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at

devlian [24]

Force acting during collision is internal so momentum is conserve so (initial momentum = final momentum) in both directions Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west. Let Vx is and Vy are final velocities of car in +x and +y direction respectively. initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)

- 750*25 + 1150*0 = (750+1150)
Vx initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)

750*0 - 1150*5 = (750+1150)
Vy from here you can calculate Vx and Vy so final velocity V is

<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>)
</span>
and angle make from +ve x axis is

<span>θ=<span>tan<span>−1</span></span>(<span><span>Vy</span><span>Vx</span></span>)

</span><span> kinetic energy loss in the collision = final KE - initial KE</span>

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3 years ago

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A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?

Vesna [10]

Answer:

<h2>12 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

$a = \frac{6000}{500} = \frac{60}{5} = 12 \\$

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

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2 years ago