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1 year ago

5

A student measures the speed of a rolling ball three times. She adds the measurements and divides by 3.What quantity did the stu

dent calculate? average speed varying speed constant speed instantaneous speed

2 answers:

Vinvika [58]1 year ago

5 0

Hey there!

A ball rolls 3 times. If she adds the three measurements and divides it by 3, it means that this is the average speed. The average speed is found by adding all measurements and dividing by the amount of numbers.

I hope this helps!

professor190 [17]1 year ago

5 0

Average Speed is correct

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A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is

Step2247 [10]

Answer:

The decelerating force is $3\times 10^{- 11}\ N$

Solution:

As per the question:

Frontal Area, A = $10\ m^{2}$

Speed of the spaceship, v = $1\times 10^{6}\ m/s$

Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

A = Area

v = velocity

t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

4 0

1 year ago

Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para

Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

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1 year ago

Unlike acceleration and velocity, speed does not need to specify

frosja888 [35]

Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.

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1 year ago

Alejandro has collected over 35 studies focusing on the impact of vitamin C on a person's cold duration. He hopes to combine the

Misha Larkins [42]

Since Alejandro hopes to combine the findings from the collected studies and draw a general conclusion regarding the subject, he is using statistical tool of inference.

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Descriptive statistical tools
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More on types of statistics can be found here: brainly.com/question/13335435

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2 years ago

A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0

Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

$vsin(\alpha) + gt = 0$

$t = \frac{-vsin(\alpha)}{g}$

So the vertical distance it travels within time t is

$y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}$

$y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}$

$y = \frac{-v^2sin^2(\alpha)}{2g}$

Similarly the horizontal distance it travels within time t is:

$x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}$

$x = \frac{-v^2sin(2\alpha)}{2g}$

We can pre-calcualte the quantity $\frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54$

So $y = 57.54sin^2(\alpha)$

$x = 57.54sin(2\alpha)$

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

8 0

1 year ago