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tia_tia [17]
3 years ago
10

How would you go about measuring the speed of a vehicle? What measurements would you have to take? What calculations would you h

ave to perform?
Physics
1 answer:
White raven [17]3 years ago
3 0

Answer:

For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. If the velocity is non-uniform all you can say is what the average speed is.

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Heights of men on a baseball team have a​ bell-shaped distribution with a mean of 166 cm 166 cm and a standard deviation of 5 cm
kaheart [24]

Answer:

95 %

99.7 %

Explanation:

\mu = 166 cm = Mean

\sigma = 5 cm =  Standard deviation

a) 156 cm and 176 cm

166-5\times 2=156

166+5\times 2=176

From the empirical rule 95% of all values are within 2 standard deviation of the mean, so about 95% of men are between 156 cm and 176 cm.

b) 151 cm and 181 cm

166-5\times 3 =151

166+5\times 3=181

The empirical rule tells us that about 99.7% of all values are within 3 standard deviations of the mean, so about 99.7% of men are between 151 cm and 181 cm.

3 0
3 years ago
An electronics technician wishes to construct a parallelplate capacitor using rutile (k=100) as the dielectric. The area of the
My name is Ann [436]

Given k=100, Area if plate A = 1.00 cm2

    = 1 x (10-2)2 m^{2}

Thickness of rutile ds = t = 1.00 mm

      = 1 x 10-3 cm

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd

When a dielectric is placed between capacitors plates then capacitance c1 = KoAd

Here d = t = 1 x 10-1 cm, o = 8.854 x 10-12 c^{2} m^{-2}  N

Now c1 = 100 X 8.859 X 10^{-12}  X 1 X (10-2)^{21}  X 10^{-2}

= 8.859 X 10-1910-3= 8.85 X 10-11 F= 8.85 X 10-12 F

 C1  = 88.5 X 10-12 F

 C1  = 88.5  PF

Correct Option: A

What purposes serve parallel plate capacitors?

The following are some uses for parallel plate capacitors:

  • This kind of capacitor is utilized in rechargeable energy systems such as batteries.
  • These capacitors are used in systems for dynamic digital memory.
  • Such capacitors are used in pulsed laser and radar circuits.

To learn more about Parallel plate capacitors, visit:

brainly.com/question/12733413

#SPJ4

5 0
2 years ago
Explain which planets have seasons and why some planets have none at all.
Sergeu [11.5K]

Answer:

Some planets have seasons some don't bc of the distance from the sun some of them are too cold or too hot to have seasons

8 0
3 years ago
What are the units for density and when do you use each
Delicious77 [7]

Every unit of density is

(a unit of mass) divided by (a unit of volume) .

The one most widely used is [ gram/cubic centimeter ] , but there's no reason that you MUST use this same unit every time you talk about density. You can use (any unit of mass) divided by (any unit of volume) that you want ... the best choices are always the ones that end up with the most convenient number.

Whatever units of mass and volume you decide to use, it's easy for anybody to convert it to [ gram/cubic centimeter ] if they want to.

8 0
4 years ago
Let us suppose the magnitude of the original Coulomb force between the two charged spheres is FF. In this scenario, a third sphe
Vinil7 [7]

Answer:

F ’= 1/32 F

We see that the value of the force is the initial force over 32

Explanation:

In this problem the sphere that is touching the others is connected to ground, after each touch,

Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half

½ (Q / 2) = ¼ Q

For the red dial repeat the same scheme

with the first touch the charge is reduced to Q / 2

with the second touch e reduce to ½ (Q / 2) = ¼ Q

with the third toce it is reduced to ½ (¼ Q) = ⅛ Q

Now let's analyze what happens to the electric force

if the force is F for when the charge of each sphere is Q

        F = k Q Q / r²

with the remaining charge strength is

        F ’= k (¼ Q) (⅛ Q) / r²

        F ’= 1/32 k Q Q / r²

        F ’= 1/32 F

We see that the value of the force is the initial force over 32

6 0
3 years ago
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