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3 years ago

As important as it is to plan ahead,sometimes you

Physics

2 answers:

Natali5045456 [20]3 years ago

5 0

Answer:

B. Cant stop things from going wrong.

Explanation:

To me it's the only reasonable answer...

fiasKO [112]3 years ago

3 0

Can’t stop from things going a different way than how you plan it to be.

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A 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up

bekas [8.4K]

The speed of the first car just before the collision is 26.73 km/h.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given a 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up and slide together with a speed of 15.0 km/h.

Second car is parked, so its velocity will be zero.

808 x u +632 x 0 = (808 +632 ) x 15

u = 26.73 km/h

Thus, the speed of the first car just before the collision is 26.73 km/h

Learn more about conservation of momentum principle

brainly.com/question/14033058

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6 0

2 years ago

What are some sources of error in a refraction lab using pins and a glass block?

ser-zykov [4K]

DONT WORRY

ANSWER KO YAN NG MADALI

PERO BRAINLIEST MUNA

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6 0

2 years ago

The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th

pickupchik [31]

Answer:

The focal length of the lens in water is $f_{water} = 262.68 cm$

The focal length of the mirror in water is $f =79.0cm$

Explanation:

From the question we are told that

The index of refraction of the lens material = $n_2$

The index of refraction of the medium surrounding the lens = $n_1$

The lens maker's formula is mathematically represented as

$\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

Where $f$ is the focal length

$n$ is the index of refraction

$R_1 and R_2$ are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air we have

$\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

When immersed in liquid the formula becomes

$\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]$

The ratio of the focal length of the the two medium is mathematically evaluated as

$\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }$

From the question

$f_{air }$ = 79.0 cm

$n_2 = 1.55$

and the refractive index of water(material surrounding the lens) has a constant value of $n_1 = 1.33$

$\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}$

$f_{water} = 262.68 cm$

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

7 0

3 years ago

A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:

34kurt

Answer:

a) $v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

b) $v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

c) $v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

d) $v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

Explanation:

For this case we know that the velocity is $v = 200 cm/s = 2m/s$

$v_f$ represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

$v = \sqrt{\frac{T}{\rho}}$

And the linear density is defined as:

$\rho = \frac{m}{L}$

And if we replace this we got:

$v = \sqrt{\frac{TL}{m}}$

If the tension mass is doubled we have this:

$v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s$

The velocity increase by a factor of $\sqrt{2}$

Part b

If we mass is quadrupled we have this:

$v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s$

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

$v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s$

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

$v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s$

The velocity not changes.

7 0

3 years ago

Any good and bad insulators of thermal energy, hot or cold, and why?

Free_Kalibri [48]

It really all depends weather the temperature

7 0

4 years ago