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

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

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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

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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:

$\omega_f = 3.584\ rad/s$