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3 years ago

5

A skier skis down a slope with a constant acceleration of 3 m/s?. If she reaches the bottom in 10 seconds, how long is the slope

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1 answer:

Hoochie [10]3 years ago

4 0

Answer:

in a town a population increased from 2000 to 214 2 years in a town with the population increase from 1018 during the same time period in which town is person to increase population

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Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

Leno4ka [110]

(a) 3.58 km 45° south of east

The total displacement is given by:

$d=vt$

where

v is the average velocity

t is the time

The average velocity is:

v = 3.53 m/s

While we need to convert the time from minutes to seconds:

$t=169 min \cdot 60 s/min = 10140 s$

Therefore, the magnitude of the displacement is

$d=(3.53)(10140)=35794 m = 3.58 km$

And the direction is the same as the velocity, therefore 45° south of east.

(b) 5.53 m/s 90° south of east

The velocity of the air relative to the ground is

$v_a = 2.00 m/s$

and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is

v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed ( $v_a$ ). Since these two vectors are in opposite direction, we have

$v= v'-v_a$

Therefore we find v', Allen's velocity relative to the air:

$v'=v+v_a = 3.53 + 2.00 = 5.53 m/s$

The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

(c) 56.1 km at 90° south of east.

Since Allen's velocity relative to the air is

v' = 5.53 m/s

Then the displacement of Allen relative to the air will be given by

$d'=v't$

and substituting,

$d'=(5.53)(10140)=56074 m = 56.1 km$

And the direction is the same as that of the velocity, therefore will be 90° south of east.

3 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

$y = \frac{m\lambda}{d}$

Here,

m = Number of order bright fringe

$\lambda$ = Wavelength

d = Distance between slits

Both distance are the same, then

$y_1 = y_2$

$\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}$

$\frac{m_1\lambda_1}{m_2\lambda_2} =1$

Rearranging to find the second wavelength

$m_1 \lambda_1 = m_2 \lambda _2$

$\lambda_2 = \frac{m_1\lambda_1}{m_2}$

$\lambda_2 = \frac{7(629)}{8}$

$\lambda_2 = 550.3nm$

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

7 0

2 years ago

A spindle connects two wheels with fixed axles by a fan belt. If the effort wheel is larger than the resistance wheel, which axl

MariettaO [177]

Power = Iω (constant) as they are connected together, since effort axle has large radius than resistance axle, so moment of inertia of effort axle is also more as compared to resistance axle, so angular speed of effort axle is less than the resistance axle. So answer is B. resistance axle will have more angular speed as its moment of inertia is less for the same power.

8 0

2 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

3 0

3 years ago

Which of the following accurately describe some aspect of gravitational waves? Select all that apply.

steposvetlana [31]

<h2>Answers:</h2>

-The first direct detection of gravitational waves came in 2015

-The existence of gravitational waves is predicted by Einstein's general theory of relativity

-Gravitational waves carry energy away from their sources of emission

<h2>Explanation:</h2>

Gravitational waves were discovered (theoretically) by Albert Einstein in 1916 and "observed" for the first time in direct form in 2015 (although the results were published in 2016).

These gravitational waves are fluctuations or disturbances of space-time produced by a massive accelerated body, modifying the distances and the dimensions of objects in an imperceptible way.

In this context, an excellent example is the system of two neutron stars that orbit high speeds, producing a deformation that propagates like a wave,<u> in the same way as when a stone is thrown into the water</u>. So, in this sense, gravitational waves carry energy away from their sources .

Therefore, the correct options are D, E and F.

5 0

3 years ago