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3 years ago

According to the Big Bang Theory, the universe is __________________. A) constant B) depleting C) expanding D) exploding

Physics

2 answers:

DiKsa [7]3 years ago

6 0

C) expanding hope this answer helps

natta225 [31]3 years ago

5 0

C.) <span>According to the Big Bang Theory, the universe is "Expanding"

Hope this helps!</span>

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The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

6 0

3 years ago

10.0 ppm 10.4 ppm 10.2 ppm 10.8 ppm 10.1 ppm 10.5 ppm 10.1 ppm Using the new instrument on the first day the technicians got a v

iren [92.7K]

Answer:

The datapoint 9.0 ppm is outlier at the 90% confidence level.

Explanation:

The old data has following values

mean=10.5 mm

standard deviation 0.2 mm

Now the mean of new values is calculated as following

$mean=\frac{10+10.4+10.2+10.8+10.1+10.5+10.1}{7}\\mean=10.3 ppm$

So the value as 9.0 ppm can be considered easily as outlier in this regard.

3 0

3 years ago

a 70 kg skydiver opens her parachute. The force due to air resistance is now 1200 N. what is the acceleration of the skydiver

Natasha_Volkova [10]

Around 50 I think so

7 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

If the elevation in reservoir b is 100m, what must the elevation in reservoir a be if thevolume flow rate through the cast-iron

Zinaida [17]

The elevation in reservoir at the rate of flow using is 03m/s is 114m.

The Reynolds range is the ratio of inertial forces to viscous forces. The Reynolds variety is a dimensionless variety used to categorize the fluids structures in which the impact of viscosity is crucial in controlling the velocities or the flow sample of a fluid.

The reason of the Reynolds number is to get a few experience of the relationship in fluid glide between inertial forces (this is those that maintain going by using Newton's first law – an item in motion stays in movement) and viscous forces, this is people who cause the fluid to come back to a forestall because of the viscosity of the fluid.

calculation,

Let L = 100 m pipe

L1 = 150 m pipe

H f = friction losses

Using Reynolds number, relative roughness, friction co- effiicients and friction losses

Substitute the value in equation

Z = 110= 0.48= 3.54

Z = 114m

Therefore water surface elevation at reservoir is 114 meter.

Learn more about rate of flow here:-brainly.com/question/21630019

#SPJ4

6 0

2 years ago