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2 years ago

14

Did you mention the correct independent variable – the substance’s properties (such as solubility, conductivity, and state of ma

tter)?
yes
no
somewhat

Did you mention the correct dependent variable – the type of bond, ionic or covalent?
yes
no
somewhat

2 answers:

goblinko [34]2 years ago

3 0

Answer:

1. Somewhat 2.Yes

Explanation:

stiv31 [10]2 years ago

3 0

Answer:

C,a

Explanation:

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John runs around a 126.5 m circular track 3.5 times in 4.17 minutes. What is his average speed?

sdas [7]

Average speed = distance traveled / time

average speed = (126.5 m * 3.5 laps) / (4.17 min)

= 106.2 m/min

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3 years ago

ludmilkaskok [199]

Answer:

1) ironing a shirt 2) writing on surfaces 3) working of an eraser

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2 years ago

Help please :pensive:

tino4ka555 [31]

Answer:

0m/s²

Explanation:

Given parameters:

Initial velocity of the boat = 8m/s

Final velocity = 8m/s

Time taken = 4s

Unknown:

Acceleration of the boat = ?

Solution:

Acceleration is the rate of change of velocity with time.

It is mathematically expressed as;

A = $\frac{v - u}{t}$

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

Insert the parameters and solve;

A = $\frac{8-8}{4}$ = 0m/s²

6 0

2 years ago

If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will

densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

$P=\frac{V^2}{R}$

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

$P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P$

so, the power dissipated will quadruple.

6 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago