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vredina [299]
1 year ago
6

If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many images wo

uld you see in the viewer?
Physics
1 answer:
sammy [17]1 year ago
6 0

Explanation:

The question was incomplete. Here is the complete question.

If you punched thousands of holes in the aluminium foil of the scope, how many images would you see in the viewer.

a) Hundreds of small images

b) a few bright images

c) one large blurry image

d) no images at all.

So the correct option is c) one large blurry image.

You might be interested in
What is the weight of a 1-kilogram brick resting on a table?
MakcuM [25]

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

The weight of an object is calculated as mass of the object × its acceleration due to gravity

W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

= 10kgm/s² or 10Newtons

5 0
1 year ago
A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
1 year ago
The weight of a bucket is 186 N. The bucket is being raised by two ropes. The free-body diagram shows the forces acting on the b
amm1812
Fnet=(115+106)-186= 34 N

mass=Force/g= 186N/9.8m/s^2 = 18.98 kg

a=fnet/mass => 34N/18.98kg = 1.79 m/s^2

so A= 1.8m/s^2
4 0
2 years ago
Which is an example of a positive feedback loop? i. warmer temperatures on earth's surface decrease the evaporation of water. ii
sergejj [24]
It would be "W<span>armer temperatures on earth's surface decrease the evaporation of water."</span>
6 0
2 years ago
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
1 year ago
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