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Solnce55 [7]
4 years ago
14

For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen

ts are given in the stemplot below. The median point total for this class.

Physics
1 answer:
Talja [164]4 years ago
5 0

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

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An object with a mass M = 250 g is on a plane inclined at 30º above the horizontal and is attached by a string to a mass m = 150
malfutka [58]

Answer:

0.495 m/s

Explanation:

T = tension force in the string connecting the two objects

M = Mass of the object on inclined plane = 250 g = 0.250 kg

m = Mass of the hanging object = 150 g = 0.150 kg

a = acceleration of each object

From the force diagram, force equation for the motion of the object on the inclined plane is given as

T - Mg Sin30 = Ma\\T = Mg Sin30 + Ma

From the force diagram, force equation for the motion of the hanging object on the inclined plane is given as

mg - T = ma\\T = mg - ma

Using the above two equations

Mg Sin30 + Ma = mg - ma

(0.250)(9.8) Sin30 + (0.250) a = (0.150) (9.8) - (0.150)a

a = 0.6125 ms^{-2}

h = height dropped by the hanging object = 10 cm = 0.10 m

v = Speed gained by the object

Speed gained by the object can be given as

v = sqrt(2ah)\\v = sqrt(2(0.6125)(0.20))\\v = 0.495 ms^{-1}

5 0
3 years ago
Can someone help me with one through seven I will mark you the brainly
ki77a [65]

Answer:

1.  F = M x A

2. Force  

3. 2nd Law: Force

4. a, b, c (in order)

5. 3rd Law: Action and Reaction

6. b, c, a (in order)

7. 1st Law: Inertia

3 0
3 years ago
Read 2 more answers
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 12,900 kg. The thrust of its engines is
Alexandra [31]

Answer:

Acceleration a=0.5 m/s²

Explanation:

Given data

Mass m=12,900 kg=1.29×10⁴kg

Thrust of engine F=28,000 N=2.8×10⁴N

gravitational acceleration g=1.67 m/s²

To find

Acceleration

Solution

As we know  that

W_{weigth}=mg\\ W=(1.29*10^{4}kg )(1.67m/s^{2} )\\W=21543N

The net force can be given as

F_{net}=F_{thrust}-W\\F_{net}=(2.8*10^{4}-21543)N\\   F_{net}=6457 N

From Newtons second law of motion we know that

F=ma\\a=F/m\\a=\frac{6457N}{12,900kg}\\ a=0.5m/s^{2}

6 0
3 years ago
Where is the sun's energy most concentrated
Maru [420]

B. At the equator

Explanation:

The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.

The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator (0^{\circ}), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.

5 0
3 years ago
A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past t
marta [7]

Answer:

Explanation:

Given that

The window height is 2m

And the window is 7.5m from the ground

Then the total height of the window from the ground is 7.5+2=9.5m

It takes the ball 0.32sec travelled pass the window.

When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')

Now using the equation of free fall during this window travels

S=ut-½gt² against motion.

S=2, g=9.81, t=0.32sec

Then,

S=u't-½gt²

2=u'×0.32-½×9.81×0.32²

2=0.32u'-0.5023

2+0.5032=0.32u'

Then, 0.32u'=2.5032

u'=2.5032/0.32

u'=7.82m/s

This is the initial velocity as the ball got the the window

Now, let analyse from the window bottom to the ground which is a distance of 7.5m

Using the equation of free fall again

v²=u²-2gH

In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,

While u is the original initial velocity from the throw of the ball

Then,

u'²=u²-2gH

7.82²=u²-2×9.81×7.5

61.146=u²-147.15

61.146+147.15=u²

Then, u²=208.296

So, u=√208.296

u=14.43m/s

The initial velocity of the ball form the throw is 14.43m/s

6 0
3 years ago
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