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4 years ago

Which describes a scientist being creative?

Physics

2 answers:

AleksandrR [38]4 years ago

4 0

Answer: Sara tries turning a test tube upside down to collect a gas.

A scientist is considered to be creative when he approaches a problem with new different ways. The conventional way is to design an experiment and take detailed notes, reading referenced studies previously done. Sara tries turning a test tube upside down to collect a gas is a creative way as Sara tries something different.

Veronika [31]4 years ago

3 0

The best answer I believe Sara. By letting the test tube sit upright she is not collecting anything as gas vapors rise. By thinking outside of the box and turning it upside down she can then collect the gas in the test tube.

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A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

Nastasia [14]

Answer:

750 nm

Explanation:

$d$ = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

$D$ = screen distance = 4.8 m

$y$ = position of first bright fringe = $\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m$

$n$ = order = 1

Position of first bright fringe is given as

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

$0.002 = \frac{(1)(4.8)\lambda }{0.0018}$

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

3 0

3 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago

The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?

kipiarov [429]

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e I = $\frac{Q}{t}$ ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

Q = I × t

= 1.5 × 2

= 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

5 0

3 years ago

From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:

Oduvanchick [21]

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m

also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

3 0

3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

4 years ago