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3 years ago

What determines what will be tested in a scientific experiment

1 answer:

5 0

What you want to test and your hypothesis does.
For example, say you came up with a hypothesis that 'The higher the temperature, the higher the reaction rate will be'.
Your independent variable (the one you change) will be temperature. The dependent variable (the one that changes because of the independent) will be the reaction rate (e.g. bubbles produced).

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

lf an object's velocity changes from 25 meters per second to 15 meters per second in 2.0 seconds, the magnitude of the object's

Mila [183]

Answer:

-5 m/s^2

Explanation:

$acceleration = \frac{change \: in \: velocity}{time} \\ \\ = \frac{15 - 25}{2} \\ \\ = \frac{ - 10}{2} \\ \\ = - 5 \: m {s}^{ - 2} \\ \\ negative \: sign \: represents \: that \: \\ object \: is \: slowing \: down.$

6 0

3 years ago

Read 2 more answers

Describe how a bowling pin will react to the unbalanced force of a bowling ball.

Jlenok [28]

Because the forces are not balanced the pin will fall down and go backwards and the force of the bowling ball will keep going after acting upon the non active pin

7 0

3 years ago

Read 2 more answers

A 6.80 $\mu C$ particle moves through a region of space where an electric field of magnitude 1230 N/C points in the positive $x$

Murljashka [212]

Answer:

v = -227.785 m/s

Explanation:

The electric field exerts the following force on the electric particle:

F = qE

$F = 6.65 \times 10^{-6} \times 1230$

$F = 0.0081795 \ N$

The magnetic field exerts the following force on the particle::

F = qvB

$F = 6.65\times 10^{-6} \times v \times 1.32$

$F = 8.778 \times 10^{-6} \times v$

Total force acting is:

F = qvB + qE

$6.18 \times 10^{-3} = 0.0081795 + 8.778 \times 10^{-6} \times v$

$v = \dfrac{6.18 \times 10^{-3} -0.0081795 }{8.778 \times 10^{-6}}$

v = -227.785 m/s

5 0

3 years ago

Two planets are orbiting a star in a distant galaxy the first has a semimajor axis of 150 x 10^6 km an eccentricity of 0. 20 and

e-lub [12.9K]

For two planets orbiting a star in a distant galaxy, having the first semimajor axis of 150 x 10^6 km, the period of the second plane is mathematically given as

T=2.2earth year

<h3>What is the period of the second planet?</h3>

Generally, the equation for the time period is mathematically given as

$T^{2} \appox a^{3}$

Therefore, for planet one

$1^2\approx (150 x 10^6)^3$

In conclusion, For planet 2

T^2=(250/150)^3

T=√4.632

T=2.2earth year