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IgorLugansk [536]
3 years ago
11

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

Physics
1 answer:
monitta3 years ago
3 0

Answer:

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

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Suppose you hit a 0.058-kg tennis ball so that the ball then moves with an acceleration of 10 m/s2. If you were to hit a basketb
stiks02 [169]

Answer:

1 m/s²

Explanation:

Force = mass × acceleration

F = ma ............ Equation 1

Where F = force, m = mass, a = acceleration.

Given: m = 0.058 kg, a = 10 m/s²

Substitute into equation 1

F = 0.058(10)

F = 0.58 N.

If the same force was used to hit the baseball,

F = m'a

a = F/m'.............. Equation 2

Where M' = mass of the baseball.

Given: F = 0.58 N, m' = 0.58 kg.

Substitute into equation 2

a = 0.58/0.58

a = 1 m/s²

8 0
3 years ago
Read 2 more answers
Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
koban [17]

Answer:

A

   N  = 1340.86 \ slits  / cm

B

    \theta  = 15.7^o

Explanation:

From the question we are told that  

      The wavelength is  \lambda  =  650 \  nm  =  650  *10^{-9} \  m  

        The angle of  first bright fringe is  \theta  =  5^o  

        The order of the fringe considered is  n  =1

Generally the condition for constructive interference is  

       dsin (\theta ) = n * \lambda

=>    d =  \frac{1 *  650 *10^{-9 }}{ sin(5)}

=>    d = 7.458 *10^{-6} \  m

Converting to cm

           d = 7.458 *10^{-6} \  m = 7.458 *10^{-6}  * 100 =  0.0007458 \  cm

Generally the number of grating pre centimeter is  mathematically represented as

           N  =  \frac{1}{d}

=>         N  =  \frac{1}{0.0007458}

=>         N  = 1340.86 \ slits  / cm

Considering question B  

   From the question we are told that

     The first wavelength is  \lambda_1 =  650 \ nm  =  650 *10^{-9} \  m

     The second wavelength is  \lambda_2 = 429 \  m   =   420 *10^{-9 } \  m

      The order of the fringe is  n  =  2

       The grating is  N =  5000 \  slits / cm

Generally the slit width is mathematically represented as

              d =  \frac{1}{N  }

=>          d =  \frac{1}{ 5000  }

=>          d =   0.0002 \  c m  =  2.0 *10^{-6} \ m

Generally the condition for constructive interference for the first ray is mathematically represented as

         d sin(\theta_1) =  n *  \lambda_1

=>      \theta_1 = sin^{-1} [\frac{ 2 *  \lambda }{d}]

=>       \theta_1 = sin^{-1} [\frac{ 2 *   650 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_1 = 40.5 ^o

Generally the condition for constructive interference for the second ray is mathematically represented as

         d sin(\theta_2) =  n *  \lambda_2

=>      \theta_2 = sin^{-1} [\frac{ 2 *  \lambda_1 }{d}]

=>       \theta_2 = sin^{-1} [\frac{ 2 *   420 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_2 = 24.8  ^o

Generally the angular separation is mathematically represented as

            \theta  =  \theta_1 - \theta_1

=>          \theta  = 42.5^o -  24.8^o

=>          \theta  = 15.7^o

4 0
3 years ago
A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h
BartSMP [9]
Specific Gravity of the fluid = 1.25 
Height h = 28 in
 Atmospheric Pressure = 12.7 psia
 Density of water = 62.4 lbm/ft^3 at 32F
 Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
 Difference in pressure = 1.26 psia
 (a) Pressure in the arm that is at Higher 
 P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
 (b) Pressure in the tank that is at Lower
 P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
4 0
3 years ago
A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest
34kurt

Displacement = (straight-line distance between the start point and end point) .

Since the road east is perpendicular to the road north,
the car drove two legs of a right triangle, and the magnitude
of its final displacement is the hypotenuse of the triangle.

    Length of the hypotenuse = √ (215² + 45²)

                                              =  √ (46,225 + 2,025)

                                              =   √ 48,250

                                              =       219.7 miles .

8 0
3 years ago
Read 2 more answers
A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th
Usimov [2.4K]

Answer:

\omega_f = 3.584\ rad/s

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed  = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

\theta = \omega_i t + \dfrac{1}{2}\alpha t^2

5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2

\alpha= \dfrac{10}{2.8^2}

       α = 1.28 rad/s²

now, calculation of angular velocity

\omega_f = \omega_i + \alpha t

\omega_f =0 +1.28\times 2.8

\omega_f = 3.584\ rad/s

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0
3 years ago
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