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3 years ago

What is happening in the first part of this picture?

Physics

1 answer:

d1i1m1o1n [39]3 years ago

6 0

Answer:

kenitic energy

Explanation:

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Flying against the wind, an airplane travels 2670 km in 3 hours. Flying with the wind, the same plane travels 11,070 km in 9 hou

xxMikexx [17]

Answer:

speed of plane in still air = 1060 km/h

speed of wind = 170 km/h

Explanation:

Let teh speed of plane in still air is vp and the speed of air is va.

Irt travels 2670 km in 3 hours against the wind

So,

vp - va = 2670 / 3 = 890 km/h ..... (1)

It travels 11070 km in 9 hours along the wind.

vp + va = 11070 / 9 = 1230 km/h .... (2)

Adding both the equations

2 vp = 2120

vp = 1060 km/h

and va = 1230 - vp = 1230 - 1060 = 170 km/h

5 0

3 years ago

A 55-kg woman is wearing high heels.

Grace [21]

Answer:

Pressure, $P=1.90\times 10^7\ Pa$

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{mg}{\pi r^2}$

$P=\dfrac{55\times 9.8}{\pi (0.003)^2}$

$P=1.90\times 10^7\ Pa$

So, the pressure exerted on the floor is $1.90\times 10^7\ Pa$ . Hence, this is the required solution.

7 0

3 years ago

The critical angle for a special type of glass in air is 30.8 ◦ . the index of refraction for water is 1.33. what is the critica

Alina [70]

When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted light, and all the light is reflected. The value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n2 and n1 are the refractive indices of the second and first medium, respectively.

In the first part of the problem, light moves from glass to air ( $n_a=1.00$ ) and the critical angle is $\theta_c = 30.8^{\circ}$ . This means that we can find the refractive index of glass by re-arranging the previous formula:
$n_g=n_1 = \frac{n_2}{\sin \theta_c}= \frac{1.00}{\sin 30.8^{\circ}}=1.95$

Now the glass is put into water, whose refractive index is $n_w = 1.33$ . If light moves from glass to water, the new critical angle will be
$\theta_c = \arcsin ( \frac{n_2}{n_1} )=\arcsin( \frac{n_w}{n_g} )=\arcsin( \frac{1.33}{1.95} )=\arcsin(0.68)=43.0^{\circ}$

6 0

3 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$

7 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

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