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DerKrebs [107]
3 years ago
5

Eating breakfast in the morning increases the ability to learn in school.

Physics
1 answer:
bonufazy [111]3 years ago
4 0

Answer:

Independent Variable - eating reakfast

Dependent Variable - ability to learn

Constant Variable - going to school

Explanation:

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I need help please I will give points
stira [4]

Answer:

-5 N of force

Explanation:

Hope it helps!

3 0
3 years ago
A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n
RSB [31]

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

3 0
2 years ago
*please refer to photo*
just olya [345]

Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.

<h3>What is average velocity?</h3>

An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.

<h3>What is an average?</h3>

An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.

<h3>How to calculate the average velocity?</h3>

Mathematically, the average velocity for this data set would be calculated by using this formula:

Average = [F(v)]/n

Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5

Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:

  • v₁ = 100 m/s
  • v₂ = 150 m/s
  • v₃ = 200 m/s
  • v₄ = 250 m/s
  • v₅ = 300 m/s

Substituting the parameters into the formula, we have:

Vavg = [300 + 450 + 500 + 250 + 300)/5

Vavg = 1800/5

Vavg = 360 m/s.

Next, we would calculate the percent difference by using this formula:

Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100

Percent difference = [360 - 343]/360 × 100

Percent difference = 17/360 × 100

Percent difference = 0.0472 × 100

Percent difference = 4.72%.

Read more on average here: brainly.com/question/9550536

#SPJ1

3 0
1 year ago
The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the
Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

V_d = \frac{I}{nAq}

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

I = 25 A

A= 1.2*20 *10^{-6} m^2

q= 1.6*10^{-19}C

N = 8.47*10^{19} mm^{-3}

V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}

V_d = 7.68*10^{-5}m/s

The hall voltage is given by

V=\frac{IB}{ned}

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}

V = 3.84*10^{-6}V

5 0
3 years ago
A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m
olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

v = v_0 -gt

Here,

v = Final velocity

v_0 = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

-30 = v_0 -9.8(4)

v_0 = 9.2m/s

Now the position can be calculated as,

y = h +v_0t -\frac{1}{2}gt^2

When it has the ground, y=0 and the time is t=4s,

0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2

h = 41.6m

Therefore the cliff was initially to 41.6m from the ground

7 0
3 years ago
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