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2 years ago

7

A projectile was launched horizontally with a velocity of 468m/s, 1.86m above the ground how long did it take the projectile to

hit the ground

1 answer:

BartSMP [9]2 years ago

5 0

Answer:

0.616 s to 3 significant figures

Explanation:

Using the equation s=ut+0.5at^2 in the vertical direction:

1.86=0×t+0.5gt^2

1.86=4.905t^2

t^2=1.86÷4.905

t=0.6157961457

=0.616 s (3sf)

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A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m Ã 3.00 m Ã 5.00 m. 1. what is the

cupoosta [38]

1) The volume of the raft is the product between the lenghts of its three dimensions:
$V = (2.00 m)(3.00m)(5.00m)=30 m^3$

2) The mass of the raft is the product between its density, d, and its volume, V:
$m=dV=(650 kg/m^3)(30 m^3)=19500 kg$

3) The weight of the raft is the product between its mass m and the gravitational acceleration, $g=9.81 m/s^2$ :
$W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N$

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
$W_a = W- \rho_W V_{disp} g$
where $\rho _W = 1000 kg/m^3$ is the water density and $V_{disp}$ is the volume of displaced fluid.
The density of the raft ( $650 kg/m^3$ ) is smaller than the water density ( $1000 kg/m^3$ ), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
$W=B=\rho _W V_{disp} g$
and therefore, the apparent weight will be zero:
$W_a = W-B=W-W=0$

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
$B=\rho_W V_{disp} g$
When the raft is completely immersed in the water, the volume of fluid displaced $V_{disp}$ is equal to the volume of the raft, $V_{disp}=V$ . Therefore the buoyancy in this situation is
$B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N$
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
$B=W=1.91 \cdot 10^5 N$

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
$mg = m_{disp} g$
where $m_{disp}= \rho_W V_{disp}$ is the mass of the displaced water. From the previous equation, we obtain that $m_{disp}=m=19500 kg$ .

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship $m=dV$ we can rewrite $m=m_{disp}$ as:
$d V =d_W V_{disp}$
and so
$V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3$

8) The volume of water displaced is (point 7) $19.5 m^3$ . This volume is now "filled" with part of the volume of the raft, therefore $19.5 m^3$ is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, $30 m^3$ :
$\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%$
Viceversa, the volume of raft above the water level is $30 m^3-19.5 m^3 = 10.5 m^3$ . Therefore, the fraction of volume of the raft above water level is
$\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%$

9) Let's repeat steps 5-8 replacing $\rho _W$ , the water density, with $\rho_E=806 kg/m^3$ , the ethanol density.

9-5) The buoyant force is given by:
$B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N$
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
$m_E = m = 19500 kg$

9-7) Using the relationship $d= \frac{m}{V}$ , we can find the volume of displaced ethanol:
$V_E = \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3$

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: $24.2 m^3$ . Therefore, the fraction of raft's volume below the ethanol level is
$\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%$
Consequently, the raft's volume above the ethanol level is
$30 m^3 - 24.2 m^3 = 5.8 m^3$
and the fraction of volume above the ethanol level is
$\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%$

8 0

2 years ago

Describe the distribution of wdiff in terms of its center, shape, and spread, including any plots you use

melisa1 [442]

Answer:

Figure attached

We can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.

Explanation:

We can use the following R code to obtain the data for wdiff:

source("http://www.openintro.org/stat/data/cdc.R") #obtain the info

nrow(cdc) # number of elements

names(cdc) # obtain the name for the variable

[1] "genhlth" "exerany" "hlthplan" "smoke100" "height" "weight" "wtdesire" "age"

[9] "gender"

wdiff represent the difference between desired weight (wtdesire) and current weight (weight) and we can obtain the data with the following code:

wdiff <- (cdc$weight-cdc$wtdesire)

And now we can create the histogram with this code

hist(wdiff,xlim =c(-100,150))

> mean(wdiff)

[1] 14.5891

> median(wdiff)

[1] 10

And the result is on the figure attached.

And we can conclude that majority of the values are positive. And we can say that is skewed to the right because the Median< Mean is and we have most of the values at the left of the distribution.

3 0

2 years ago

For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?

Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

$85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}$

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity ( $v_t$ ), in order to obtain the angular velocity of the wheel:

$v_t=w*R\\w=\frac{v_t}{R}$

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

$w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}$

And now, recalling that $2\pi$ radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by $2\pi$ :

rotational frequency = $\frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}$

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2 years ago

Write a hypothesis about how the force applied to a cart affects the acceleration of the cart. Use the "if . . . then . . .becau

Lesechka [4]

"If one increases the force on an object, its acceleration increases too because the push it feels is greater"

We have the 2nd law of Newton that relates the 3 concepts; F=m*a. We have that if the mass of an object increases (put weight in luggage), the accelearation decreases; in fact it is inversely proportional to the mass. Hence if the mass is doubled, acceleration is halved. Accelerations is proportional to force; if one doubles the force, the acceleration doubles too.

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2 years ago

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Shareen performs a skit to model a method of charging. In the skit, a painter shakes her hand and gets paint on her.

Veseljchak [2.6K]

Answer:

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

Explanation:

Conduction phenomenon of charging is the process of charging in which two bodies are made in contact with each other so that charges are transferred due to potential difference of two bodies.

here we know that when hands are shake then it will have paint on it. so here due to hand shake the hands are in contact with charge particles and due to contact the electrons are transferred to the hand.

Now here we need to assume that charge of paint must be opposite that of the charge on the hand because only due to opposite charge attraction the paint must be transferred to the hand

SO here correct answer will be

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

5 0

2 years ago

Read 2 more answers