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2 years ago

Which refers to an object’s resistance to any change in its motion? force acceleration gravity inertia

2 answers:

7 0

<em>Inertia</em> is the property of all matter by which it tends to remain in constant, uniform motion until it's acted on by an external force.

wlad13 [49]2 years ago

4 0

Answer:

Inertia

Explanation:

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A student measured the maximum mass of salt that can dissolve in 100mL of water at five different temperatures. Which variable s

Wewaii [24]

Answer: The answer is B, The variable that should go on the x- axis is the temperature of water.

Explanation: In an experiment, we must have two variables; the independent variable and the dependent variable.

If the results of the experiment is reported in graphical format, the independent variable is plotted on the x axis while the dependent variable is plotted on the y axis.

In this case, the independent variable is the temperature of the water while the dependent variable is the mass of salt dissolved in 100 mL of water.

Therefore, the variable that should go on the x- axis is the temperature of water.

3 0

1 year ago

One strategy in a snowball fight is to throw

faltersainse [42]

Answers:

a) $\theta_{2}=23\°$

b) $t=1.199 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=11.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=67\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))$ (9)

$x=9.043 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=45.99\°$

$\theta_{2}=22.99\° \approx 23\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(11.1 m/s)sin(67\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.085 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(11.1 m/s)sin(23\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=0.885 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.085 s - 0.885 s$

Finally:

$t=1.199 s$

7 0

2 years ago

Which wave causes the medium to vibrate only in a direction not parallel to the waves motion?

Paha777 [63]

The correct answer to the question is : Transverse wave.

EXPLANATION :

Before going to answer this question, first we have to understand the longitudinal and transverse wave.

LONGITUDINAL WAVE : A longitudinal wave is a mechanical wave in which the direction of vibration of particles is parallel to the direction of wave propagation. It moves in the form of compression and rarefaction.

For instance, sound wave.

TRANSVERSE WAVE : A transverse wave is a mechanical wave in which the direction of vibration of particles is perpendicular to the direction of wave propagation. It moves in the form of crests and troughs.

For instance, the wave created in a pond when a stone is dropped into it.

Hence, the correct answer of this question is transverse wave.

4 0

2 years ago

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Monarch butterflies can sustain air speeds of 4.8 m/s while migrating. A certain monarch wants to fly to a location due east 4.7

AnnyKZ [126]

It will take the butterfly 2.3 km to reach the destination

6 0

2 years ago

Read 2 more answers

can you help me create a sketch of two different objects with one that has a greater density than the other?

sergey [27]

I don't know how good you are at sketching ... I'm terrible.
But you can put the point across in a dramatic way if you
can sketch a bowling ball and a basketball ... you'll need
to clearly identify them with the markings you sketch on
each ball.

They're the same shape and nearly the same size, but
there's a huge difference in their densities.

6 0

2 years ago