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3 years ago

The answer please it’s very simple this is 7th grade science

Physics

2 answers:

olganol [36]3 years ago

7 0

Answer:

Kinetic Energy

Explanation:

The explanation is the cart is moving and the people are not so the answer is kinetic energy.

allochka39001 [22]3 years ago

3 0

The greatest point for kinetic is at the bottom and in the middle it is in half and at the top it is at the highest in potential energy.

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Two students were climbing stairs at school. Student 1 has a weight of 700 N. Student 2 has a weight of 650 N. How much power wo

KIM [24]

Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

From the question,

We are to calculate the power each student would have to climb the flight of stairs.

Power can be calculated using the formula

$P = \frac{F \times d}{t}$

Where

P is Power

F is the force

d is the distance

and t is the time

NOTE: The weight of the students represent the force

For student 1

F = 700 N

d = 4 m

t = 6 s

∴ $P = \frac{700 \times 4}{6}$

P = 467 W

For student 2

F = 650 N

d = 4 m

t = 6 s

∴ $P = \frac{650 \times 4}{6}$

P = 433 W

Hence, Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

Learn more here: brainly.com/question/18801566

3 0

2 years ago

Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it

jek_recluse [69]

Answer:

a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

I = I₁ cos² θ'

we substitute

I = (I₀ cos² tea) cos² (θ - 90)

cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

I = Io cos² θ sin² θ

1= cos²θ+ sin²θ

sin²θ = 1 - cos²θ

I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

$\frac{dI}{d \theta}$ = 0

\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

cos θ - 2 cos³ θ = 0

cos θ ( 1 - 2 cos² θ) = 0

The zeros of this function are in

θ = 90º

1-2cos²θ =0 cos θ = 0.25 θ = 75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0

3 years ago

A car on a freeway speeds up to get around another car. The car speeds up from 20 m/s to 35 m/s in 5 seconds.

Tanzania [10]

Answer:

Initial speed = 20 m/s

Final speed = 35 m/s

Time to speed up = 5 seconds

Explanation:

Directly from the information given:

Initial speed = 20 m/s

Final speed = 35 m/s

Time to speed up = 5 seconds

5 0

3 years ago

What causes seasons to occur?

LUCKY_DIMON [66]

It’s C I hope it helps you

3 0

3 years ago

Read 2 more answers

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

yaroslaw [1]

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

$E_{new} = E/4$

4 0

3 years ago