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3 years ago

Someone please help me will give BRAILIEST!!!!!

Physics

1 answer:

ratelena [41]3 years ago

4 0

The speed increases, because as the angle increases (the wing slants up more steeply), the air has to go farther to get over the wing.

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A light bulb emitting 1750 Im of luminous flux shines on a book that is 3.0

denis-greek [22]

The illumination of the book is found to be approximately equal to 15 lx.

Option 1.

Explanation:

Luminous flux is the measure of power of the light emitted from a source and illumination is the property of the light to illuminate a given surface when it falls on the surface. So the measure of the region the light flux illuminates the surface is termed as illumination.

As here the luminous flux is given as 1759 lm and the distance of the object which has to be illuminated is placed 3 m away from the source ,the illumination is found to be directly proportional to the luminous flux and inversely proportional to the square of the distance between them

$illumination = \frac{Luminous flux}{4\pi d^{2} } \\\\illumination = \frac{1750}{4*3.14*3*3} = 15.48 lx = 15 lx$

So, the illumination of the book is found to be approximately equal to 15 lx.

5 0

3 years ago

1. Explain one way gravity affects objects.

Alex Ar [27]

Answer:

it makes it fall to the ground instead of float

Explanation:

8 0

3 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

3 years ago

An open container holds ice of mass 0.500 kgkg at a temperature of -17.4 âCâC . The mass of the container can be ignored. Heat i

MaRussiya [10]

Answer:

a) t = 235.2minute

b) 260.94minutes

Explanation:

Heat energy is defined as the energy required to change the temperature of a substance by 1kelvin.

It is expressed as H = mc∆t where m is the mass of the water

c is the specific heat capacity of the water/ice

∆t is the change in temperature

Total heat required to melt the ice

H = mLice + mc∆t

Lice is the latent heat of fusion of ice

a) To calculate how much time tmelts passes before the ice starts to melt, we will only calculate heat energy absorb by the ice before it melts.

H = mLice

H = 0.50(334000)

H = 167,000Joules

If heat is supplied to the container at the constant rate of 710 J/minute, the time taken before the ice starts to melt will be:

t = 167,000/710

t = 235.2minutes

b) To calculate how much time trise does it take before the temperature begins to rise above 0°C, we will calculate the total energy absorbed at 0°C first.

H = 0.50(334000) + 0.50(2100)(0-(-17.4)

Note that the ice melts at 0°C which will be the final temperature

H = 167,000+18270

H = 185,270oules

Since heat is supplied to the container at the constant rate of 710 J/minute

710Joules = 1minute

185,270Joules= x

x = 185,270/710

x = 260.94minutes

The time taken before the temperature begins to rise is 260.94minutes:

6 0

3 years ago

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Give one example of a question that science would not be able to test.

Strike441 [17]

DOES GOD EXIST? That is one of the best.

8 0

4 years ago

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