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2 years ago

A fruit bat falls from the roof of a cave. We know that her potential energy was

Physics

1 answer:

bulgar [2K]2 years ago

5 0

Answer:

v = 15.65 m/s

Explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

Pi = m g h = 600 J

and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

m = 600 J / (g 12.5) = 4.9 kg

Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

with Pf = 0 (just touching the ground)

KEf= 1/2 4.9 (v^2)

and we solve for the velocity:

600 J = 0 + 1/2 4.9 (v^2)

v^2 = 600 * 2 / 4.9 = 244.9

v = 15.65 m/s

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Which of the following choices best represents a potential stressor?

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an unexpected visitor as it begins making you feel uncomfortable but it's get calm

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Select the correct answer. An object has one force acting on it. It is a 33-newton force pointing downward. To create a net forc

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2 years ago

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A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th

s344n2d4d5 [400]

Answer:

15 cm

Explanation:

$h_{o}$ = Diameter of the coin = 15 mm

$h_{i}$ = Diameter of the image of coin = 5 mm

$d_{o}$ = distance of the coin from mirror = 15 cm

$d_{i}$ = distance of the image of coin from mirror = ?

Using the equation

$\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}$

$\frac{d_{i}}{15} = \frac{- (5)}{15}$

$d_{i}$ = - 5 cm

$R$ = radius of curvature

Using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}$

$\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}$

$R$ = - 15 cm

6 0

3 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

4 0

3 years ago

A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev

Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²

a= 4×π²×2×3²

a=710.6 m/s²

5 0

3 years ago