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Ulleksa [173]
3 years ago
5

4.7 cm. The tube is sealed at one end and loaded with lead shot to give it a total mass of M = 130 g. The tube floats in water (

of density 1 g/cm3 ) in vertical position, loaded end down. What is the depth of the bottom end of the tube? Answer in units of cm.
Physics
1 answer:
babunello [35]3 years ago
6 0

Answer:

Depth is 7.49cm

Explanation:

Given that;

The diameter of tube is (D)=4.7cm

Area of the tube is =πD^2/4

Area is 3.142*4.7*4.7/4=17.35cm2

Mass(M)=130g

Density =1g/cm^3

Recall that Density =mass /volume

Hence volume =130/1=130cm^3

Also note

Volume=area*depth

130=17.35*depth

Depth=130/17.35=7.49cm

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Ety ratio
horrorfan [7]

3) The work done is D. zero

4) The kinetic energy is B. 180 J

5) The potential energy is A. 120 J

6) The work done depends on B. position

7) The example of non-renewable energy is C. coal

8) The power expended is 3\cdot 10^4 W

9) The efficiency is A. 100%

10) The velocity ratio is 5

Explanation:

3)

The work done by a force acting an object is given by:

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

When the force is applied perpendicular to the direction of motion,

\theta=90^{\circ}

Therefore, the work done is:

W=Fd(cos 90^{\circ})=0

4)

The kinetic energy of a body is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the girl in this problem, we have

m = 40 kg

v = 3 m/s

Therefore her kinetic energy is

K=\frac{1}{2}(40)(3)^2=180 J

5)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g=10 m/s^2 is the acceleration of gravity

h is the heigth of the object relative to the ground

For the ball in this problem,

m = 0.4 kg

h = 30 m

So, the potential energy is

PE=(0.4)(10)(30)=120 J

6)

A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.

Gravitational and electric fields are examples of conservative fields. In fact:

  • When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height \Delta h between the two points, not on the path taken during the fall
  • When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field

For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.

7.

  • Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
  • Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.

Therefore, the example of non-renewable energy in this case is

C. Coal

8.

For an object pushed by a force F and moving at a constant velocity v, the power expended is given by

P=Fv

where F is the force and v is the velocity.

for the rocket in this problem, we have:

F = 10 N is the force propelling the rocket

v = 3000 m/s is its velocity

Substituting into the equation, we find the power expended:

P=(10)(3000)=30,000 W = 3\cdot 10^4 W

9.

The efficiency of a machine is given by

\eta = \frac{W_{out}}{W_{in}}

where

W_{in} is the energy in input to the machine

W_{out} is the useful work in output from the machine

For a real machine, the useful work in output is always lower than the energy input, because part of the energy is "wasted" and converted into thermal energy due to the presence of internal frictions. However, for an ideal machine, all the input energy is converted into useful work, so

W_{out}=W_{in}

And therefore the efficiency is

\eta=1

which means 100%.

10.

The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.

VR=\frac{d_{eff}}{d_{load}}

In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.

For the system in the problem, there are 5 pulleys: therefore, this means that when the effort moves 5 metres, the load moves 1 metres, therefore the velocity ratio is

VR=\frac{5}{1}=5

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

5 0
4 years ago
A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed
galina1969 [7]

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

3 0
3 years ago
A force can exist between two charged particles or objects even if they're as small as subatomic particles. Between which of the
ASHA 777 [7]
If both particles have the SAME electrical charge, then they repel.
If they have DIFFERENT electrical charge, then they attract.

Protons have  +  charge .
Electrons have  -  charge .

So two protons (A) or two electrons (D) push apart.

One proton and one electron (C) pull together.
3 0
3 years ago
Read 2 more answers
PLEASE HELP ME WILL GIVE BRAINLIEST
weqwewe [10]

Answer:

A.always changing

Explanation:

7 0
3 years ago
Read 2 more answers
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0
3 years ago
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