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Mars2501 [29]
3 years ago
13

State one activity in our daily life by using the characteristic of iron rod.​

Physics
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

In our house ceilings.

Explanation:

In our daily life, iron rod is used in the construction of ceiling of our homes which made our ceiling stronger enough to make it multi story. It is extensively used in big construction projects such as multi story buildings and other tall buildings etc. So we can say that iron rod is used for construction of big houses and buildings.

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suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str
tatyana61 [14]

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

5 0
3 years ago
If the new moon happens on January 15th, what shape will it be on February 6th?
Jobisdone [24]

-- From January 15 to February 6 is a period of 22 days.

-- The period of the full cycle of moon phases is 29.53 days.

-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.

-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at <em>Third Quarter</em>.  That's the <em>left half of a disk </em>(viewed from the northern hemisphere).

3 0
3 years ago
a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
4 years ago
Name two kinds of longitudinal waves and explain how you know they are longitudinal
jasenka [17]
First of all Longitudinal waves is a matter in the medium that moves parallel to the direction of the wave travels. 
1st example: sound travels parallel.
2nd example: when you talk you will here your voice again. because all the frequency bounce back to you.

In the other hand Transverse wave matter in the medium moves perpendicular to direction the wave travels.
For example: light is a good example of transverse wave.
6 0
3 years ago
A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied
Juliette [100K]

Answer:

v = 50.5 m/s

Explanation:

F = (m)(^v/^t)

115N = (0.04551kg)(v/(0.020s))

2,526.917161 m/s² = v/(0.020s)

v = 50.53834322 m/s

v = 50.5 m/s

8 0
3 years ago
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