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3 years ago

Anyways chile , what y’all doing virtual or school ?

Physics

2 answers:

Lady bird [3.3K]3 years ago

8 0

First year of college entirely virtual, r.i.p.

goldenfox [79]3 years ago

3 0

Answer:bggfd

this cow be doing that viurtuaExplanation:

sedioisiij

You might be interested in

If a positively charged body is moved against an electric field it will gain?

Gnoma [55]

Answer:

positive charge

Explanation:

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative.

hope it helps :)

please mark it the brainliest!

5 0

4 years ago

It is 4.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

Musya8 [376]

Answer:

Explanation:

To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.

But first we have to calculate the time

$t_{1}=\frac{x}{v_{1}}=\frac{4km}{10\frac{km}{h}}=0.4h=0.4(3600s)=1440s\\t_{2}=\frac{x}{v_{2}}=\frac{4km}{3\frac{km}{h}}=1.3h=1.3(3600s)=4800s\\$

Now we use E=W*t for both times

$E_{1}=t_{1}W_{1}=(1440s)(700W)=1008000J\\E_{2}=t_{2}W_{2}=(4800s)(290W)=1392000J\\$

A. Hence, by running the energy consumed is lower.

E1=1008000J

E2=1392000J

C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.

5 0

3 years ago

If acceleration of a particle at any time is given by a=2t+5 the velocity after 5 seconds, if it starts from rest is?

kow [346]

The velocity after 5 seconds is 50 m/s.

3 0

3 years ago

Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

Tamiku [17]

Answer:

$7.0\cdot 10^{-13}C$

Explanation:

The magnitude of the electrostatic force between two charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

$r=0.070 cm =7\cdot 10^{-4} m$ is the distance between the charges

$q_1=q_2=q$ since the charges are identical

$F=9.0\cdot 10^{-9}N$ is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

$F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C$

8 0

3 years ago

A girl and a boy pull in opposite directions on strings attached to each end of a spring balance. Each child exerts a force of 2

BabaBlast [244]

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO generally: (left or down) would be considered negative in an equation. And the other person (right or up) would be considered positive. So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

0 is the number of equilibrium.

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

thanks for coming to my TED talk. I hope they don't delete this answer.

5 0

3 years ago