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igor_vitrenko [27]
4 years ago
12

Kate gathered three boxes of the same size made of different materials: glass, clear plastic, and aluminum painted black. She pl

aced them on a window sill in the sun for an hour and then measured the warmth of the air in each box. In this experiment, the warmth of the air is a(n)
Independent variable
Dependent variable
Constant
Control
Physics
2 answers:
Umnica [9.8K]4 years ago
8 0

<u>Answer:</u> The correct answer is dependent variable.

<u>Explanation:</u>

Independent variable is defined as the variable which remain as such and has no effect on the change of another variable. For Example: Time

Dependent variable is defined as the variable whose value changes with respect to any other variable.

Constant variable is defined as the variable whose value cannot be change if any value is assigned to it.

According to the question:

Three boxes of different materials having same size is kept in the Sunlight to check the warmth of air inside the box.

Thus, sunlight is independent variable and warmth of air inside the box is the dependent one because if the sunlight falling on the boxes increases, the temperature of the air also increases and vice-versa.

Hence, the correct answer is dependent variable.

Thepotemich [5.8K]4 years ago
6 0

Answer:

Dependent variable.

Explanation:

Based on the description, the experiment is designed to confirm or reject a hypothesis about a heat absorption patterns by  hollow objects with different transparency. Temperature appears to be the main target variable of interest - this is the <em>dependent</em> variable. Material that each box is made of (specifically, its transparency) is in this case the independent variable, which is determined by the experimenter.

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Your car is traveling behind a jeep. Both are moving at the same speed, so the velocity of the jeep relative to you is zero. A s
7nadin3 [17]

Answer:

No.

Explanation:

The spare tire is also moving at the speed of the jeep. When it separates from the jeep, both will be moving at the same horizontal speed, only that the spare tire will also fall at the same time. This if we neglect air friction, that could be a problem since it could cancel the horizontal velocity of the spare tire, although the jeep is blocking most of the air for the spare tire to feel any significant effect in such a short time.

6 0
3 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
4 years ago
hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward
dimulka [17.4K]

For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

t = 5 m / 6.65 m/s

t = 0.75 s

Result:

The time will be <u>0.75 seconds.</u>

7 0
3 years ago
Read 2 more answers
Suppose there is a bright fringe at P Would this bright fringe move closer to O, move further away, or be unchanged, t 1.5 marks
fiasKO [112]

Answer:

Explanation:

The distance of a fringe from centre is proportional to wavelength of light

and inversely proportional to separation of slits. The expression for distance x is given by

x = nλ D / d

where λ is wave length , D is screen distance and d is slit separation.

So first option only is correct because

1 ) the wavelength of blue light is less than that of red

2) Intensity of light does not affect distance of fringe from the centre.

3.

Diffraction symbolises bending of light around sharp edges like slits or boundaries of opaque objects etc.Due to this reason , we do not observe sharp boundary of shadow of an object. Instead around the boundary of shadow, we observe bands of bright and dark color which are also called fringes.

The phenomena of diffraction is explained by wave theory of light.

3 0
3 years ago
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
4 years ago
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