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3 years ago

Define newtons first law of motion with examples(class 9 ) plssss itz urgent

2 answers:

6 0

Newton's first law of motion states that an object maintains its state motion or that of rest unless subjected to some external force.

It means that if an object is moving then it will not stop until we apply force on it. Similarly, if an object is at rest, then it will not move until we apply force on it. Objects also maintain their speed, it means that their speed will not change unless we apply force.

This is because objects have inertia. Inertia is the measure of the property of the object to oppose any change in its motion. More the mass more the inertia. This is why this law is also known as the law of inertia.

Example:

1. A bag kept on the top of bus, falls backward because it maintains its state of rest and bus starts moving.

Akimi4 [234]3 years ago

5 0

<h3>Newtons first law state that if a body is in motion it will be in motion and if a body is in rest it will tend to be in rest. This phenomena is also called INERTIA. Example: We tend to fall sideways when a car turn suddenly</h3>

You might be interested in

If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa

vovikov84 [41]

Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,

F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.

7 0

3 years ago

. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp

MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0

2 years ago

Dumbo the elephant weights 13,000 pounds. What force would the Earths surface apply to Dumbo? 2.2 pounds = 1 kg

snow_lady [41]

Answer:

57908 N

Explanation:

Let's first convert Dumbo's mass into kg using the given relationship: 2.2 pounds =1 kg.

Then, 13000 lbs = 13000/2.2 kg = 5909 kg

Now, let's find the force of gravity on Dumbo at the surface of the earth, which would be in magnitude equal to the normal force that the Earth's surface applies on Dumbo.

F = m * a = 5909 kg * 9.8 m/s^2 = 57908 N

8 0

2 years ago

Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil

tatyana61 [14]

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$ (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$

4 0

3 years ago

A stunt driver drives a car horizontally off the edge of a cliff at 3.8m/s and reaches the water below 2.5s later.

andreyandreev [35.5K]

A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance

Here is my work, I find making charts like this one to find knowns and unknowns can be helpful

4 0

3 years ago