1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sladkaya [172]
4 years ago
14

Can you change how much power you use while exercising?

Physics
1 answer:
Nina [5.8K]4 years ago
8 0

Answer:Most treadmills and stationary bikes use electricity, but what if you could produce electricity while exercising? With a pedal-power generator, you can! And you can use the electricity immediately to power a television, computer, stereo or other electronics — or store it in batteries to use it later.

Explanation:

You might be interested in
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
Plz answer this very soon
tester [92]

Answer:

Im gonna say it is answer A:) Hope this helps!

Explanation:

6 0
3 years ago
Read 2 more answers
In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor
VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596 kg-m2

I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2

by perpendicular axis theorem

I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2

4 0
3 years ago
All of the following were barriers to minority participation in early psychology except
Masja [62]
<span>All of the following were barriers to minority participation in early psychology except school desegregation. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that you were looking for and the answer has come to your great help.</span>
4 0
3 years ago
Read 2 more answers
g Show that the first derivative of the magnitude of the net magnetic field of the coils (dB/dx) vanishes at the mid- point P re
never [62]

Answer:

Hi Carter,

The complete answer along with the explanation is shown below.

I hope it will clear your query

Pls rate me brainliest bro

Explanation:

The magnitude of the magnetic field on the axis of a circular loop, a distance z  from the loop center, is given by Eq.:

B = NμοiR² / 2(R²+Z²)³÷²

where

R is the radius of the loop

N is the number of turns

i is the current.

Both of the loops in the problem have the same radius, the same number of turns,  and carry the same current. The currents are in the same sense, and the fields they  produce are in the same direction in the region between them. We place the origin  at the center of the left-hand loop and let x be the coordinate of a point on the axis  between the loops. To calculate the field of the left-hand loop, we set z = x in the  equation above. The chosen point on the axis is a distance s – x from the center of  the right-hand loop. To calculate the field it produces, we put z = s – x in the  equation above. The total field at the point is therefore

B = NμοiR²/2 [1/ 2(R²+x²)³÷²   + 1/ 2(R²+x²-2sx+s²)³÷²]

Its derivative with respect to x is

dB /dx=  - NμοiR²/2 [3x/ (R²+x²)⁵÷²   + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]

When this is evaluated for x = s/2 (the midpoint between the loops) the result is

dB /dx=  - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷²   - 3(s/2)/(R²+s²/4)⁵÷² ] =0

independent of the value of s.

8 0
3 years ago
Other questions:
  • What can pressure be measured by?
    7·1 answer
  • Which of these forces are acting within a nucleus? Check all that apply.
    8·1 answer
  • Sr-90 has a half-life of T1/2 = 2.85 a (years). How much Sr-90 will remain in a 5.00 g sample after 5.00 a? Show all of your wor
    15·1 answer
  • Sics and Chemistry A (2nd Six Weeks) - SC3200SFA/01
    14·1 answer
  • Q 1: Calculate the pressure.
    10·1 answer
  • Help please physics !!
    9·1 answer
  • An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
    5·1 answer
  • Kbhgyugybhhv vihiklmb
    8·1 answer
  • Select the correct answer
    11·2 answers
  • How has technology changed farming?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!