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3 years ago

Renald completed only one trial of his experiment. what effect will this most likely have?

1 answer:

6 0

Remember the rule of thumb that every person learns in Elementary Science? You must do multiple experiments in order to get to a conclusion. In order for a conclusion to be valid you must test the conclusion multiple times. You wouldn't want a doctor to just test an aspirin 1 time on 1 patient and say yes it works correct? No, you would want him/her to test on multiple patients in multiple settings and conditions so that when you take an aspirin you know that it will work for what you are taking it for. So ..... with all that being said.....Your answer is (A). The results are more likely to have errors.

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Pls help me l will make it brainlest

nikklg [1K]

Answer:

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

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=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

⇒ α = <u>0.02</u>

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Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

⇒ β = 2 × α

⇒ β = 2 × 0.02

⇒ β = <u>0.04 °C⁻¹</u>

3 0

2 years ago

Read 2 more answers

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

Why are renewable sources of energy better than nonrenewable sources?

Anton [14]

Because it does not produce waste, thus it doesn't harm the environment. also renewable sources are infinite.

5 0

3 years ago

Read 2 more answers

A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e

prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

$F_E$ is the electric force =?