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3 years ago

How much force is needed to accelerate a 4.0-kg physics book to an acceleration of 2.0 m/s2?

Physics

1 answer:

Verdich [7]3 years ago

4 0

F=MA
F=(4 kg)(2 m/s^2)
F=8 N
Hope this helps

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Which particle is a negatively charged ion?

wel

Answer:

option D is correct

Explanation:

cl has 17 electrons if it has 18 it becomes cl-

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2 years ago

A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when

Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

$v^2-u^2=2ad$

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

$a=-8.0 m/s^2$ is the acceleration

Solving for u, we find the initial velocity:

$u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s$

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3 years ago

Help pls thx!!!!!!! <br> this is my last question i hope!

irakobra [83]

Answer:

Left

Explanation:

newtons is a measure of force. Since there is more newtons(force) pushing to the left, the object will move left. the 15 newtons cancel each other out, leaving only 5 newtons pushing to one side with no resistance.

4 0

2 years ago

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A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.

Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 65.0° )

${\theta_r}$ is the angle of refraction ( 53.0° )

${n_r}$ is the refractive index of the refraction medium (unknown liquid, n=?)

${n_i}$ is the refractive index of the incidence medium (oil, n=1.38)

Hence,

$1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}$

Solving for ${n_r}$ ,

Refractive index of unknown liquid = 1.56

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3 years ago

How long will it take to travel 200 m traveling at 10 m/s? Follow example below.

polet [3.4K]

Answer:

$variables - d = 200m \: v = 10 m {s}^{ - 1} \\ equation \: - v = \frac{d}{t} \\ 10 = \frac{200}{t} \\ cross \: multiply \\ 10t = 200 \\ \frac{10t}{10} = \frac{200}{10} \\ t = 20s$

It will take 10 seconds to travel 200m at a speed of 10m/s

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

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2 years ago

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