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stealth61 [152]

1 year ago

8

You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

happen to have around. With the lens close to your eye and the animal at the lens's focal point, what angular magnification M M do you achieve? Assume your near point is at 25.0 cm.

1 answer:

Zepler [3.9K]1 year ago

4 0

Answer:

The angular magnification is $M = 2.808$

Explanation:

From the question we are told

The focal length is $f = 8.9cm$

The near point is $n_p = 25.0cm$

The angular magnification is mathematically represented as

$M = \frac{n_p}{f}$

Substituting values

$M = \frac{25}{8.9}$

$= 2.808$

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Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They

denis-greek [22]

Answer:

v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

Em₀ = U = m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

m g h = ½ m v²

v² = 2gh

v = √ 2gh

Let's calculate

v = √ (2 9.8 150)

v = 54.2 m / s

3 0

1 year ago

A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0

2 years ago

A 1 kg coconut falls vertically from a height of 10 meters. Use the principle of conservation of mechanical

Kay [80]

Answer:

Explanation:

Question 1:

Mass=1kg

Acceleration due to gravity=9.8m/s^2

Height=10m

on the before falling it has potential energy

Potential energy=mass x acceleration due to gravity x height

Potential energy=1 x 9.8 x 10

Potential energy=98 joules

Question 2:

Potential energy=kinetic energy base base on energy transformation

Kinetic energy=(mass x (velocity)^2)➗2

98=(1 x(velocity))^2 ➗ 2

Cross multiplying

98 x 2=(velocity)^2

196=(velocity)^2

Velocity=√(196)

Velocity=14

Velocity=14m/s

5 0

2 years ago

A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops

Anit [1.1K]

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

4 0

1 year ago

Read 2 more answers

A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t

Grace [21]

Answer: $25.97 rad/s^2$

Explanation:

Given

radius of wheel $r=0.28 m$

mass of wheel $m=8.80 kg$

Force $F=32 N$

Moment of Inertia of solid wheel $I=\frac{mr^2}{2}$

$I=\frac{8.8\times 0.28^2}{2}$

$I=0.344 kg-m^2$

Torque is given by

$\tau =F\times r=I\times \alpha$

$32\times 0.28=0.344\times \alpha$

$\alpha =25.97 rad/s^2$

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

3 0

2 years ago