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3 years ago

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Suppose a certain scale is not calibrated correctly, and as a result, the mass of any object is displayed as 0.75 kilogram less

than its actual mass. What is the correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale?

1 answer:

Papessa [141]3 years ago

7 0

Answer

A certain scale is not calibrated correctly,

Mass displayed as 0.75 kilogram less than its actual mass.

The scale of display will be equal to '1'

The correlation between actual mass and displayed mass is 1 because correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.

So, the correct answer will be "1"

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Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago

The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is

Ghella [55]

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

8 0

3 years ago

The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh

stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, $P_{headlight}$ = 38 W

The power at which the kick starter operates in parallel, $P_{kick \ starter}$ = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/ $P_{headlight}$ = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ $P_{kick \ starter}$ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

$P_{headlight}$ = I² × R₁

∴ $P_{headlight}$ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, $P_{headlight, S}$ ≈ 0.134 W

$P_{kick \ starter}$ = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, $P_{kick \ starter, S}$ ≈ 2.123 W

The total power they will consume, $P_{Total}$ = $P_{headlight, S}$ + $P_{kick \ starter, S}$

Therefore;

$P_{Total}$ ≈ 0.134 W + 2.123 W = 2.257 W

4 0

3 years ago

A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for

Elena-2011 [213]

Answer:

1) 10.1 s 2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

In the vertical direction, as the bomb is dropped, its initial velocity is 0.
So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

$\Delta y = \frac{1}{2}*a*t^{2} (1)$

where Δy = -500 m (taking the upward direction as positive).
a=-g=-9.8 m/s²
Replacing these values in (1), and solving for t, we have:

$t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s$

The time required for the bomb to reach the earth is 10.1 s.

2)

In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

$x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.$

3)

The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

$v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s$

5)

If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
So, when the bomb hits the ground, the helicopter will be exactly over it.

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Where in the solar system (and beyond) have scientists found evidence of organic molecules?

mel-nik [20]

Answer:

Beyond our solar system, organic molecules have been found in giant clouds of dust and gas between stars and in star-forming regions.

In our solar system, besides Earth, organic molecules have been discovered on comets, in meteorites, on Saturn's moon Titan, in the plumes of water expelled from Saturn's moon Enceladus, and on Neptune's moon Triton.

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3 years ago