All categories

3 years ago

Write a description of the race using these words (all of them):

2 answers:

4 0

Hello! I would love to help but the link that you have, sent me to a video on learning how to count. I would be happy to help if you can try and post it in the comments. :))

Alenkinab [10]3 years ago

4 0

Answer:

of course

Explanation:

You might be interested in

What is acceleration, and what units do we use to measure it? *

Scrat [10]

Answer:

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

Explanation:

7 0

3 years ago

How will you separate mixture of metal and nonmetal objects

mojhsa [17]

I think it’s by using a magnet because metals are magnetic

7 0

3 years ago

A cold beer initially at 35ºF warms up to 40ºF in 3 min while sitting in a room of temperature 70ºF. How warm will the beer be i

vaieri [72.5K]

67 if it is left out for 20 minutes

5 0

3 years ago

Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$

if we solve for the magnitude, we arrive at

$F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}$

Hence the net force on one of the masses is

$F=8.0*10^{-10}N$

8 0

3 years ago

The base si unit for electric is the

nevsk [136]

The basic SI unit that has anything to do with electricity is the unit
of electric charge ... the Coulomb.

Every other unit connected with electricity is a combination of the
Coulomb and the other basic SI units ...kilogram for mass, the meter
for length, and the second for time.

6 0

3 years ago