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3 years ago

WHEN THE FORCE APPLIED ON AN ELASTIC OBJECT IS EQUAL AND OPPOSITE,THEN THIS FORCE MAY

Physics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer:

Elastic force

Explanation:

Elastic force occur when deformed or elastic object try to return to its original or natural length. When an elastic material is stretched, it resist it change in shape. The force act to return the elastic material to its original shape, it's then exert a counter force in opposit e direction.

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Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0

2 years ago

water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons

jekas [21]

Answer:

The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, $T_{(equilibrium)}$ reached is the same for both the cup and the kettle as given by the relation;

$\infty M_{(environ)} \times c_{(environ)} \times (T_2 - T_1) = m_{1} \times c_{(water)} \times (T_3 - T_2) + m_{2} \times c_{(water)} \times (T_4 - T_2)$

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

4 0

3 years ago

A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti

Pavel [41]

If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
d = 156.2 km

So, the ship will have to travel 156.2 km to the northwest direction.

8 0

3 years ago

Hi i need answers for this. Thank you!! this is also really important and is due tomorrow at 9am!!

Allushta [10]

Answer:

You're four sentences should include about how the roller coaster has the most potential energy at the top of the track, and the opposing energy, "kinetic" has the most kinetic energy when going down the hill.

Explanation:

Kinetic - In-Motion.

Potential - Gathering Energy to go into Motion.

( I'll try to answer questions to clear up confusion. )

7 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago