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3 years ago

Bohr found experimental evidence for his atomic model by studying

Physics

2 answers:

Alona [7]3 years ago

7 0

Answer:

Bohr found experimental evidence for his atomic model by studying <em><u>line spectra</u></em>.

Elden [556K]3 years ago

5 0

Answer:

Bohr found experimental evidence for his atomic model by studying line spectra.

Line Spectra is plural of a line spectrum and a line spectrum is an emission of light, sound, or other radiation, composed of a number of discrete frequencies or energies (a spectrum).

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Jim takes 45 seconds to walk 180 m north to a store . What is James velocity in meters per second?

solmaris [256]

4m/s due north of the store

Explanation:

Given parameters:

Time taken = 45s

Displacement = 180m

Unknown

Velocity = ?

Solution:

Velocity is the displacement of a body per unit time. Velocity is a vector quantity that has both magnitude and direction.

Velocity = $\frac{displacement}{time}$ = $\frac{180}{45}$

Velocity = 4m/s due north of the store

Learn more:

Velocity brainly.com/question/10883914

#learnwithBrainly

4 0

3 years ago

In 1932 Albert Dremel of Racine, Wisconsin, created his rotary tool that has come to be known as a dremel.

kramer

Answer:

3054.32618 rad/s²

-431.1989 rad/s²

29080

Explanation:

Converting angular speed to rad/s

$\omega=35000\times \frac{2\pi}{60}=3665.19142\ rad/s$

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3665.19142-0}{1.2}\\\Rightarrow a=3054.32618\ rad/s^2$

The average acceleration while speeding up is 3054.32618 rad/s²

The number of turns in the 1.2 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 3054.32618\times 1.2^2\\\Rightarrow \theta=2199.11484\ rad=\frac{2199.11484}{2\pi}=349.99\ rotations$

The number of rotations in the 1.2 seconds is 349.99

Number of rotations in the 45 seconds

$\frac{35000}{60}\times 45=26250\ rotations$

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3665.19142}{8.5}\\\Rightarrow a=-431.1989\ rad/s^2$

Average angular acceleration while slowing down -431.1989 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3665.19142^2}{2\times -431.1989}\\\Rightarrow \theta=15577.0668\ rad=\frac{15577.0668}{2\pi}\\ =2479.16718\ rotations$

Number of rotations while slowing down is 2479.16718

Total number of rotations is 349.99+26250+2479.16718 = 29079.15718 = 29080

3 0

2 years ago

Consider three capacitors C1, C2, and C3 and a battery. If

VLD [36.1K]

Answer:

Charge on C₁ = charge on all the three capacitors in series with it = 7.5 μC

Explanation:

Since the same voltage in the battery is used for the entire rundown,

From this information "only C₁ is connected to the battery, the charge on C₁ is 30.0 μC",

Q = C₁V = 30 μC

V = (30/C₁)

the series combination of C₂ and C₁ is connected across the battery, the charge on C₁ is 15.0 μC

The charge on both capacitors are the same and equal to 15 μC (because they are in series)

Q = (Ceq) V = 15 μC

(Ceq) = (15/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (15/V) μF

(Ceq) = 15 ÷ (30/C₁)

(Ceq) = 15 × (C₁/30) = 0.5 C₁

(1/Ceq) = (2/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₂)

(2/C₁) = (1/C₁) + (1/C₂)

(2/C₁) - (1/C₁) = (1/C₂)

(1/C₁) = (1/C₂)

C₁ = C₂

C₂ = C₁

C₃, C₁, and the battery are connected in series, resulting in a charge on C₁ of 10.0 μC.

The charge on both capacitors are the same and equal to 10 μC (because they are in series)

Q = (Ceq) V = 10 μC

(Ceq) = (10/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (10/V) μF

(Ceq) = 10 ÷ (30/C₁)

(Ceq) = 10 × (C₁/30) = 0.333 C₁

(1/Ceq) = (3/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₃)

(3/C₁) = (1/C₁) + (1/C₃)

(3/C₁) - (1/C₁) = (1/C₃)

(2/C₁) = (1/C₃)

C₁ = 2C₃

C₃ = (C₁/2)

C₁, C₂, and C₃ are connected in series with one another and

with the battery, what is the charge on C₁

The charge on C₁ is the same as the charge on all the capacitors and equal to Q,

Q = (Ceq) V

(1/Ceq) = (1/C₁) + (1/C₂) + (1/C₃)

Substituting for C₂ and C₃

C₂ = C₁ and C₃ = (C₁/2)

(1/C₂) = (1/C₁) and (1/C₃) = (2/C₁)

(1/Ceq) = (1/C₁) + (1/C₁) + (2/C₁)

(1/Ceq) = (4/C₁)

Ceq = (C₁/4)

Q = (Ceq) V = (C₁/4) V

But recall that V = (30/C₁) from the first connection

Q = (C₁/4) (30/C₁)

Q = (30/4) = 7.5 μC

Hope this helps!

6 0

3 years ago

Calculate the escape speed for a spacecraft from the surface of (a) mars; and from the surface of (b) jupiter. ( the escape spee

Len [333]

The escape speed for a spacecraft at the surface of a planet of mass M and radius R is:
$v= \sqrt{ \frac{2GM}{R} }$
where G is the gravitational constant. We can use this formula to solve both parts of the problem, using the data of Jupiter and Mars.

a) Mars:
Mars mass is $M=6.4 \cdot 10^{24}kg$ , while Mars radius is $R=3.4 \cdot 10^6 m$ , so the escape speed of the spacecraft at Mars surface is
$v= \sqrt{ \frac{2 (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(6.4 \cdot 10^{24} kg)}{3.4 \cdot 10^6 m} }= 5011 m/s = 5.01 km/s$

b) Jupiter:
Jupiter mass is $M=1.9 \cdot 10^{27} kg$ while its radius is $R=7.15 \cdot 10^7 m$ , so the escape speed at its surface is
$v= \sqrt{ \frac{2 (6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2})(1.9 \cdot 10^{27} kg)}{7.15 \cdot 10^7 m} }=5.95 \cdot 10^4 m/s = 59.5 km/s$

3 0

3 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current i

lisov135 [29]

Answer:

8.07 m/s, 81.7º NE.

Explanation:

The ship, due to the local ocean current, will be deviated from its original due north bearing.
In order to find the magnitude of the velocity of the ship, we need to convert a vector equation, in an algebraic one.
If we choose two axes coincident with the N-S and W-E directions, we can find the components of the velocity along these directions.
Clearly, the velocity of the ship, relative to water, is only due north, so it has no component along the W-E axis.
The local ocean current, as it is directed at an angle between both axes, has components along these axes.
These components can be found from the projections of the velocity vector along these axes, as follows:

$vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s$

The component along the N-S axis (y-axis) of the velocity of the ship will be the sum of the velocity relative to water, plus the component of the ocean current along this same axis:

$vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s$

The component along the W-E axis, is just the component of the local ocean current in this direction:

vshx = 1.17 m/s

We can find the magnitude of the velocity vector, applying the Pythagorean theorem, as follows:

$v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s$

The direction of the vector relative to the W-E axis (measured in counterclockwise direction) is given by the relative magnitude of the x and y components, as follows:

$tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg$

The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.

7 0

3 years ago