Which is found farthest from the center of an atom?

Like all planets, the planet Venus orbits the Sun in periodic motion and simultaneously spins about its axis. Just as on Earth,

Liono4ka [1.6K]

Answer:

a) F = 5.14 10⁻⁸ Hz, f = 4.76 10-8 Hz, b) v = 2.29 m / s, f = 42.5 Hz

Explanation:

a)This problem has two parts.

For the calculations relative to the planet Venus, we use that the period and the frequency are related

f = 1 / T

frequency of the orbit around the Sun

Let's reduce the period to the SI system

T = 225 days (24h / 1days) (3600 s / 1h) = 1.94 10⁷ s

F = 1 / 1.94 10⁷

F = 5.14 10⁻⁸ Hz

rotation frequency

T = 243 d = 2.1 107 s

f = 1 / T

f = 1 / 2.1 107

f = 4.76 10-8 Hz

b) give the data of some marine waves

the speed of the wave can be found with kinematics

v = x / t

v = 50.0 / 21.8

v = 2.29 m / s

If the wavelength is L = 9.28m

this distance is the distance between two consecutive ridges or valleys

λ / 2 = L

λ = 2L

λ = 2 9.28

λ = 18.56 m

the speed of the wave is

v = λ f

f = v /λ

f = 2.29 / 18.56

f = 42.5 Hz

7 0

1 year ago

One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has

vovangra [49]

Answer:

The mass of the cube is 420.8 kg.

Explanation:

Given that,

Length of edge = 38.9 cm

Density $\rho= 7.15 \times10^{3}\ kg/m^3$

We need to calculate the volume of cube

Using formula of volume

$V = 38.9^3$

$V=0.058863\ m^3$

We need to calculate the mass of the cube

Using formula of density

$\rho = \dfrac{m}{V}$

$m = V\times\rho$

$m =0.058863\times7.15 \times10^{3}$

$m=420.8\ kg$

Hence, The mass of the cube is 420.8 kg.

7 0

1 year ago

An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the essen

Anuta_ua [19.1K]

Answer:T=116.84 N

Explanation:

Given

Weight of hiker =1040 N

acceleration $a=1.1 m/s^2$

Force exerted by Rope is equal to Tension in the rope

$F_{net}=T=ma_{net}$

$T=\frac{1040}{g}\times 1.1$

$T=116.84 N$

8 0

2 years ago

The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light

Nikolay [14]

7 0

1 year ago

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What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is $W=mg=2.28 N$

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

$F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N$

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

$F_y=W=mg=2.28 N$ (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

7 0

2 years ago

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