Ask question

Ask question

All categories

3 years ago

14

A=vf-vi/t is the equation for calculating the acceleration of an object. write out the relationship shown in the equation using

words
PLEASE I NEED HELP!! I'm timed!

20 points!

1 answer:

Lelechka [254]3 years ago

3 0

Basically it is the difference in velocity divided by the time it takes to make that change.

You might be interested in

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

(c) Magnitude of the force that keeps you go around at this acceleration

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

3 years ago

a 2,000-kilogram railroad car moving at 8m/s to the right collides with a 6,000-kilogram railroad car moving at 3m/s to the west

astra-53 [7]

A freight car of mass 20,000 kg moves along a frictionless level railroad track ... After the push the skateboarder II moves with a velocity of 2 m/s to ... After the collision the cars stick to each other and ... diver jumps with a velocity of 3 m/s in opposite ... A 10 kg object moves at a constant velocity 2 m/s to the right and collides

3 0

3 years ago

What happens to an electromagnetic wave as it passes from space to matter?

alina1380 [7]

Answer:

When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected.

4 0

3 years ago

Read 2 more answers

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

3 years ago

What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai

olga nikolaevna [1]

Answer:

The magnetic flux density is $2.11\times10^{-6}\ T$

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

$B =\dfrac{\mu_{0}I}{2r}$

Where,

r = radius

I = current

Put the value into the formula

$B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}$

$B=2.11\times10^{-6}\ T$

Hence, The magnetic flux density is $2.11\times10^{-6}\ T$

3 0

3 years ago