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4 years ago

Before the car starts, the driver and all the passengers should always:

Physics

2 answers:

den301095 [7]4 years ago

6 0

Buckle and adjust their seatbelts

Evgen [1.6K]4 years ago

6 0

Answer:

buckle and adjust their seat belts

Explanation:

As we know that inertia is the property of mass due to which an object has tendency to oppose to change its state of motion.

Which we can say that object will all continue its state of motion until and unless some external force will act on it.

So here before starting the car our state of motion is REST and after starting the car we will be in motion so here due to inertia we may have change in state of motion on our body.

So in order to save our-self from change in state of motion we need to hold our body by seat belts so that we can transfer the momentum to the belt.

So here correct answer will be

buckle and adjust their seat belts

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The ball in front of you is going to drop one meter. how long will it take to fall?

Pavlova-9 [17]

The ball will take 0.45 seconds to reach the ground

<h3>How will we solve this question?</h3>

We will use Newton’s laws of motion, so

H= $(1/2)gt^2$ taking height= 1m

$1= (1/2) x 9.8 x t^2$

$t=\sqrt{10/49} = 0.45 seconds$

<h3>What are the three laws of motion given by Newton?</h3>

The 3 laws of motion given by Newton are as follows:

1) Unless an unbalanced force acts upon it, an item at rest remains at rest, and an object in motion continues to move in a straight line at a constant pace.

2) An object's acceleration is influenced by its mass and the force being applied.

3) Whenever one thing applies force to another, the second object applies the opposing, equal force to the first.

To know more about Newton's laws of motion visit:

brainly.com/question/27915475

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7 0

2 years ago

Monochromatic light of wavelength 720 nm is incident on parallel slits, of width 0.420 mm, separated by 0.560 mm. A diffraction

ki77a [65]

Answer:

1.81×10^-4Io

Explanation:

Check attachment

4 0

4 years ago

A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.

disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

a = 28.32 m/s

N = 44.4 N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = $\frac{1}{2}MR^2$

Where M is the mass and R is the radius of the wheel, then:

I = $\frac{1}{2}(8 kg)(0.3 m)^2$

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N = 0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N = 44.4 N

6 0

3 years ago

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As

Naddik [55]

Answer:

$\lambda = 1.667 nm$

$\theta = 0.8681^o$

Explanation:

From the question we are told that

The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

The is distance of the screen from the slit is $D = 2.60 \ m$

The distance between the central bright fringe and either of the adjacent bright $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally the condition for constructive interference is

$d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So $sin (\theta ) = \theta$

=> $d \theta = n \lambda$

=> $\theta = \frac{n * \lambda }{d }$

Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as

$y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

$y = D * \theta$

=> $\theta = \frac{ y}{D}$

$\frac{n * \lambda }{d } = \frac{y}{D}$

$\lambda =\frac{d * y }{n * D}$

substituting values

$\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

$\lambda = 1.667 *10^{-6}$

$\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so n= 2

Hence

$dsin (\theta ) = n \lambda$

$\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

$\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

$\theta = 0.8681^o$

7 0

4 years ago

Which motion listed below matches the graph?

mote1985 [20]

Bro I really think it might be c

7 0

3 years ago

Read 2 more answers