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
N76 [4]
3 years ago
6

PLEASE HELPP MEEE!!!!!!!!!The goal is to increase the power; therefore, it is necessary to

Physics
2 answers:
Novay_Z [31]3 years ago
4 0

Answer: The answer would be C

Explanation:

Brrunno [24]3 years ago
3 0

As we know that power is defined as rate of work done

so we will have

P = \frac{Work}{time}

so in order to increase the power as per above formula we know that either we need to increase the work or we need to decrease the time to complete that work

So here the correct answer will be

increase the work being done or decrease the time in which the work is completed.

You might be interested in
According to Newton's<br> Ist law, what will an<br> object in motion tend<br> to do?
SCORPION-xisa [38]

Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

<u>It stays in motion with the same speed and goes in the same direction.</u>

<u></u>

<em>Hope this helped </em>

<em>:)</em>

4 0
3 years ago
Read 2 more answers
Monochromatic light falling on two very narrow slits 0.048mm apart. Successive fringes on a screen 5.00m away are 6.5cm apart ne
tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}

<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>

Now, the frequency can be given as:

f = \frac{c}{\lambda}

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\

<u>f = 5.13 x 10¹⁴ Hz</u>

5 0
2 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
What is the true answer in this question getting a good night sleep ​
Alexxx [7]
Getting a good night sleep can benefit you in a lot of ways. Also that’s not a question.
4 0
2 years ago
Read 2 more answers
If someone drove for 20 minutes going 30 m/s, how far did they drive?
vladimir2022 [97]
D- 36,000m is the answer
5 0
3 years ago
Other questions:
  • Describe the results of Ernest Rutherford's gold-foil experiment and explain how his results changed ideas about the distributio
    13·1 answer
  • a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s
    9·1 answer
  • What does Hooke's law say about the relationship of a spring to the force applied to it?
    6·1 answer
  • Check all choices below that are correct. Increasing the frequency increases the current. Changing the frequency does not affect
    11·2 answers
  • The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner
    15·1 answer
  • A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strik
    5·1 answer
  • a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of
    11·2 answers
  • A body is under the action of two forces 7N and 10N. Find the resultant of the two forces if the forces are parallel and act in
    5·1 answer
  • A circuit contains two resistors in series. The voltage drop across the first is 10 V. The voltage drop across the second is als
    13·1 answer
  • A proton is projected in the positive x direction into a region of a uniform electric field →E =(-6.00 × 10⁵) i^ N/C at t=0 . Th
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!