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Keith_Richards [23]
3 years ago
13

A nature photographer is using a camera that has a lens with a focal length of 3.06 cm. The photographer is taking pictures of a

ncient trees in a forest and wants the lens to be focused on a very old tree that is 10.4 m away. a) How far must the lens be from the film in order for the resulting picture to be clearly focused? Answer in units of cm.
Physics
1 answer:
sleet_krkn [62]3 years ago
6 0

Answer:

film is at distance of 3.07 cm from lens

Explanation:

Given data

focal length = 3.06 cm

distance = 10.4 m = 1040 cm

to find out

How far must the lens

solution

we apply here lens formula that is

1/f = 1/p + 1/q

here f = 3.06 and p = 1040 so we find q

1/f = 1/p + 1/q

1/3.06 = 1/1040 + 1/q

1/ q =  0.3258

q = 3.0690 cm

so film is at distance of 3.07 cm from lens

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A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i
scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0
3 years ago
A skydiver jumps out of a plane flying at an altitude of 5,500 meters. How fast is the skydiver going when they get to the groun
andrey2020 [161]

Answer:

2,500 feet (760 meters)

Explannation: <em>At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.</em>

4 0
3 years ago
Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
vredina [299]

Answer:

The angular speed after 6s  is \omega = 1466.67s^{-1}.

Explanation:

The equation

I\alpha  = Fd

relates the moment of inertia I of a rigid body, and its angular acceleration \alpha, with the force applied F at a distance d from the axis of rotation.

In our case, the force applied is F = 22N, at a distance d = 6cm =0.06m, to a ring with the moment of inertia of I =mr^2; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I} $

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $

\alpha  = 244.44\: s^{-2}

Therefore, the angular speed \omega which is

\omega  = \alpha t

after 6 seconds is

\omega = 244.44$\: s^{-2}* 6s

\boxed{\omega = 1466.67s^{-1}}

7 0
3 years ago
A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
3 years ago
car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 k
Mice21 [21]

Answer:

32000 N

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Mass (m) of car = 400 Kg

Force (F) =?

Next, we shall determine the acceleration of the the car. This can be obtained as follow:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Acceleration (a) =?

v² = u² + 2as

0² = 40² + (2 × a × 10)

0 = 1600 + 20a

Collect like terms

0 – 1600 = 20a

–1600 = 20a

Divide both side by –1600

a = –1600 / 20

a = –80 m/s²

The negative sign indicate that the car is decelerating i.e coming to rest.

Finally, we shall determine the force needed to stop the car. This can be obtained as follow:

Mass (m) of car = 400 Kg

Acceleration (a) = –80 m/s²

Force (F) =?

F = ma

F = 400 × –80

F = – 32000 N

NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.

7 0
2 years ago
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