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IgorLugansk [536]
2 years ago
11

The name of the bright star near the center that the other stars appear to be circling is:

Physics
1 answer:
monitta2 years ago
3 0

Answer:

the sun or Me not sure but one of them has to be ...

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In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t
USPshnik [31]

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

5 0
3 years ago
A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day. the effective capacity, however, is 700
Alekssandra [29.7K]

A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day. the effective capacity, however, is 700 boards per day. recently the production facility has been producing 950 boards per day. The design capacity utilization is (950/100) *100 = 95 %

3 0
3 years ago
FOR BRAINLIEST!<br><br> Someone draw me a Rube Goldberg Machine.
tatyana61 [14]
This ain’t mine but here is someone else’s

8 0
2 years ago
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

6 0
3 years ago
9. Calculate the distance (in km) that Charlie runs if he maintains the average
Karo-lina-s [1.5K]

<u>Correct Question:</u>

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

<u>Answer:</u>

The total distance covered by Charlie is 8 km in 1 hour.

<u>Explanation:</u>

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$t=1 hour

As we know the formula to evaluate the total distance d when the average velocity and time is given;

v=\frac{d}{t}

d=v \times t

d=8 \times 1

d=8 k m

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

5 0
3 years ago
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