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BlackZzzverrR [31]
3 years ago
13

14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.

Physics
1 answer:
sineoko [7]3 years ago
3 0

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

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Memorize this and you'll be able to do ALL of these:  <em>1  kg = 1,000 g</em>

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1,000 g = 1.000 kg

500 g = 0.500 kg

100 g = 0.100 kg

50 g = 0.050 kg

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4 0
2 years ago
Plz do all plz i will give brainest and thanks to best answer do it right
lara31 [8.8K]

Answer:

I'm not really sure but I think it's choice a

5 0
2 years ago
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An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev
marishachu [46]

Answer:

Explanation:

Given that,

Force applied to pedal F = 50N

Angular velocity ω = 10rev/s

We know that, 1rev = 2πrad

Then, ω = 10rev/s = 10×2π rad/s

ω = 20π rad/s

Length of pedal r = 30cm = 0.3m

Power?

Power is given as

P = τ×ω

We need to find the torque τ

τ = r × F

Since r is perpendicular to F

Then, τ = 0.3 × 50

τ = 15 Nm

Then,

P = τ×ω

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P = 942.48 Watts

power delivered to the bicycle by the athlete is 942.48 W

6 0
3 years ago
Which statement about activation energy is true? A. It is not affected by catalysts. B. It has no effect on the rate of the reac
HACTEHA [7]
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6 0
3 years ago
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A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
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