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4 years ago

Why should you use control while during an experiment?

Physics

1 answer:

Sergio039 [100]4 years ago

8 0

To provide a greater certainty that the observed results are not by chance.

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A plane is traveling at a velocity of 90 m/s. It accelerates at a constant rate of 1.5 m/s2 until its velocity reaches 500 m/s.

katrin2010 [14]

Answer:

The distance the plane covered while it was accelerating is 80,633.3 m

Explanation:

Given;

initial velocity of the plane, u = 90 m/s

acceleration of the plane, a = 1.5 m/s²

final velocity of the plane, v = 500 m/s

The distance covered by the plane is given as;

v² = u² + 2ad

where;

d is the distance covered by the plane;

500² = 90² + 2(1.5)d

500² - 90² = 3d

241900 = 3d

d = 241900 / 3

d = 80,633.3 m

Therefore, the distance the plane covered while it was accelerating is 80,633.3 m

5 0

3 years ago

At what distance would the repulsive force between two electrons have a magnitude of 4.0 N?

borishaifa [10]

Answer:

$7.6\cdot 10^{-15} m$

Explanation:

The electrostatic force between two electrons is given by:

$F=k\frac{e^2}{r^2}$

where

$k=8.99\cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant

$e=-1.6\cdot 10^{-19}C$ is the electron charge

r is the distance between the two electrons

In this problem, we know F=4.0 N, so we can re-arrange the equation to calculate r:

$r=\sqrt{\frac{ke^2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(-1.6\cdot 10^{-19})^2}{4.0 N}}=7.6\cdot 10^{-15} m$

5 0

3 years ago

A ball is kicked at an angle . It is intended that the ball lands in 휃=44.5°the back of a moving truck which has a trunk of leng

evablogger [386]

Answer:

$V_{0min}=19.63m/s$

$V_{0max}=20.45m/s$

Explanation:

In order to solve this problem, we must start by drawing a diagram of the situation, this will help us understand the problem better so we can take the best possible approach. (See attached picture).

So next, we can see that the ball will have a parabolic trajectory, so we can split it into its x- and y-components. Let's start by analyzing the movement of the ball in the x-direction.

$x_{b}=x_{0b}+V_{0bx}t$

we know that the x-component of the velocity of the ball can be found by using the following equation:

$V_{0bx}=V_{0b}cos \theta$

So the equation can be rewritten as:

$x_{b}=x_{0b}+V_{0b}(cos \theta) t$

we can say that the ball will start its movement from the origin, so the initial displacement is zero. So the simplified equation is:

$x_{b}=V_{0b}(cos \theta) t$

So now we can analyze the vertical movement of the ball. We can describe it with the equation:

$y_{b}=y_{0b}+V_{0by}t+\frac{1}{2}at^{2}$

we can say that the ball will start its movement at the origin so we can say that the initial position of ball is equal to zero and the final position is zero as well, so the equation turns to:

$0=V_{0by}t+\frac{1}{2}at^{2}$

we can describe the vertical velocity of the ball with the following equation:

$V_{0by}=V_{0b}sin \theta$

and the acceleration as:

$a=-g=-9.81m/s^{2}$

so we can substitute them in the original equation:

$(V_{0b}sin \theta)t-\frac{1}{2}gt^{2}=0$

which can be solved for t by factoring, so we get:

$t(V_{0b}sin\theta - \frac{gt}{2})=0$

so we get:

t=0 which is the initial moment at which the ball starts moving

$V_{0b}sin\theta - \frac{gt}{2}=0$

which solves to:

$t=\frac{2V_{b0}}{g}sin \theta$

so we can substitute this into the x-movement equation:

$x_{b}=V_{0b}cos \theta(\frac{2V_{b0}}{g}sin \theta)$

which simplifies to:

$x_{b}=\frac{2V_{b0}^{2}cos\theta sin\theta}{g}$

By using trigonometric identities we can further simplify this to:

$x_{b}=\frac{V_{b0}^{2}sin(2\theta)}{g}$

We can leave that equation there by the time and start analyzing the movement of the truck. We can determine the distance between the back of the truck and the initial position of the ball with the following equation:

$x_{b}=x_{t}+d$

we know that:

$x_{t}=V_{t}t$

from the previos steps we know that the time t can be found by analyzing the y-movement of the ball, so we can use that same equation to substitute for t, so we get:

$x_{t}=V_{t}(\frac{2V_{b0}}{g}sin \theta)$

which can be substituted into the first equation:

$x_{b}=V_{t}(\frac{2V_{b0}}{g}sin \theta)+d$

and we can combine it with the equation we got for the x-movement of the ball.

$V_{t}(\frac{2V_{b0}}{g}sin \theta)+d=\frac{V_{b0}^{2}sin(2\theta)}{g}$

So we are dealing with a quadratic equation which can be rewritten like this:

$\frac{2V_{t}V_{0b}sin\theta}{g}+d=\frac{V_{0b}^{2}sin(2\theta)}{g}$

we can rewrite it in standard form so we get:

$\frac{V_{0b}^{2}sin(2\theta)}{g}-\frac{2V_{t}V_{0b}sin\theta}{g}-d=0$

So we can proceed and substitute the values we know, so we get:

$\frac{V_{0b}^{2}sin(2(44.5^{o})}{9.81}-\frac{2(11.4)V_{0b}sin(44.5^{o})}{9.81}-7.3=0$

which yields:

$0.10192V_{0b}^{2}-1.629V_{0b}-7.3=0$

which can be solved by using the quadratic formula:

$V_{b0}=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

so we get:

$V_{b0}=\frac{-(-1.629)\pm \sqrt{(-1.629)^{2}-4(0.10192)(-7.3)}}{2(0.10192)}$

which gives us two answers:

$V_{b0min}=19.63m/s$

$V_{b0min}=-3.65m/s$

Due to the nature of the problem we only take the positive answer, since

a negative answer would mean the ball is moving to the opposite side. So the answer is:

$V_{b0min}=19.63m/s$

In order to find the maximum velocity we follow the sme procedure, with the difference that in this case d=7.3m+2m=9.3m

The additional 2 meters come from the length of the trunk of the truck.

Turning the equation into:

$0.10192V_{0b}^{2}-1.629V_{0b}-9.3=0$

and changing the answers to:

$V_{b0}=\frac{-(-1.629)\pm \sqrt{(-1.629)^{2}-4(0.10192)(-9.3)}}{2(0.10192)}$

$V_{b0max}=20.45m/s$

$V_{b0min}=-4.46m/s$

so we take only the positive answer into account leaving us with the answer:

$V_{b0max}=20.45m/s$

5 0

4 years ago

A 55.6 kg ice skater is moving at 1.73 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole.

GarryVolchara [31]

Answer:

The force exerted by the rope on her arms is 273.7 N = 0.274 kN

Explanation:

Step 1: Data given

Mass of the ice skater = 55.6 kg

Velocity = 1.73 m/s

She then moves in a circle of radius 0.608 m around the pole.

Step 2:

Force exterted by the horizontal rope is the centripetal force acting on theice skater:

Fc = M*ac

⇒ with ac = v²/r

Fc = M * v²/r

Fc = 55.6 * 1.73²/0.608

Fc =273.69 N

The force exerted by the rope on her arms is 273.7 N = 0.274 kN

7 0

3 years ago

lbvjy [14]

Similarities:
-- All three classes have a fulcrum (pivot).
-- All three classes have a point where the effort force is applied.
-- All three classes have a point where the load or resistance force is applied.
-- If you can find a place to stand and a lever that's long enough,
then you can move the Earth with a 1st or 2nd Class lever.

Differences:
-- The mechanical advantage of a 1st Class lever
can be greater than 1, equal to 1, or less than 1.
-- The mechanical advantage of a 2nd Class lever is always more than 1 .
-- The mechanical advantage of a 3rd Class lever is always less than 1 .

7 0

3 years ago