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Dmitry [639]
3 years ago
12

8. A 12kg bowling ball has a velocity of 8 m/s, and is brought to

Physics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

The change in momentum experienced by the bowling ball is 96 kgm/s.

Explanation:

Given;

mass of the bowling ball, m = 12 kg

velocity of the bowling ball, v = 8 m/s

time  of motion, t = 10 s

The change in momentum experienced by the bowling ball is equal to the impulse experienced by the bowling ball.

ΔP = J = F x t

J = ft =\frac{mv}{t} \times t = mv\\\\J = 12 \times 8\\\\J = 96 \ kg.m/s

Therefore, the change in momentum experienced by the bowling ball is 96 kgm/s.

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If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
Why did she use an red infra lamp
mixas84 [53]
To act as the Sun' was accepted but if you put 'sunlight' alone it was not accepted. The examiner wanted you to state that the infra red radiation was needed to warm up the water.
4 0
3 years ago
g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe
olga_2 [115]

Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

               The spring constant is  k = 860 N/m

               The coefficient of static friction is  \mu = 0.36

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   F_y = m_b*g

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   F_x = k *x * \mu

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      x = \frac{m_b * g}{k * \mu}

substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

3 0
3 years ago
When you look at yourself in the mirror, what is the approximate angle of incidence of the light rays?
Kitty [74]
46 degrees
I hope this helps :)
3 0
3 years ago
PLS HELP ME AS QUICK AS POSSIBLE,
Levart [38]

PLS HELP ME AS QUICK AS POSSIBLE,

THANKS :)) I'm a bit confused

Can you answer 1 and 2, then confirm 3 :))))

4 0
2 years ago
Read 2 more answers
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