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3 years ago

Edwin Hubble calculated the expansion rate of the universe. On what evidence did he base his calculation?

2 answers:

7 0

Edwin Hubble calculated the expansion rate of the universe. The evidence that he base his calculation is the differences in redshift for galaxies. The answer is letter B. the red shift of galaxies was directly proportional to the distance of the galaxy from earth. It means that bodies farther away from Earth were moving away faster. The Hubble’s constant is the ratio of distance to redshift equal to 170 kilometers per second per light year of distance.

Anni [7]3 years ago

7 0

Answer: B

Explanation:

You might be interested in

Help!!! What is the kinetic energy of a 0.1kg baseball thrown at 20m/s? With work please

True [87]

Kinetic Energy = 1/2 * mass * velocity^2

Mass = 0.1 kg

Velocity = 20 m/s

Kinetic energy = 1/2 * 0.1 * 20^2

Kinetic energy = 1/2 * 0.1 * 400

Kinetic energy = 20 J

4 0

3 years ago

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

3 years ago

Read 2 more answers

How do I go about this?

Anna71 [15]

Hi there!

(a)

Recall that:
$W = F \cdot d = Fdcos\theta$

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

$W =248(56)cos(30) = 12027.36 J$

To the nearest multiple of ten:
$W_A = \boxed{12030 J}$

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
$\boxed{W_g = 0 J}$

(d)

Recall that the force of kinetic friction is given by:
$F_{f} =\mu_k mg$

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
$W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J$

In multiples of ten:
$\boxed{W_f = -3070 J}$

(e)
Simply add up the above values of work to find the net work.

$W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}}$

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
$F_{net} = F_{Ax} - F_f$

$W = F_{net} \cdot d = (F_{Ax} - F_f)$

$W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}$

5 0

2 years ago

a boy throw a ball upward from the top of a cliff 73m high.Calculate the time in which ball will fall on the ground and the velo

LekaFEV [45]

The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:

h(t)=g/2(1.5)²+14.7(1.5)

=-4.9 x 2.25 + 22.05

=11.025m

The ball then falls for 49+11.025=60.025m, which takes:

g/2t²=60.025

t²=12.25

t=3.5 secs

Total time: 1.5+3.5=5 seconds

3 0

4 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago