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UkoKoshka [18]
3 years ago
5

Please Help me!! A baseball with a mass of 0.152 kg is moving horizontally at 32.0 m/s [E], when it is struck by a bat for 0.002

00 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20° N]. a) Find the impulse experienced by the ball. (6 marks) b) Find the average net force of the ball. (2 marks)
Physics
1 answer:
shutvik [7]3 years ago
8 0

Answer:

this is not in my standard class

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If an object is accelerating, it is traveling the same distance for each time interval of its motion
Anika [276]

Answer:

false

Explanation:

if an object is accelerating the object will not travel the same distance every time interval

4 0
3 years ago
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A wire is stretched just to its breaking point by a force F . A longer wire made of the same material has the same diameter. The
ICE Princess25 [194]

Answer: No

Explanation:

The force F required is equal to the Force exerted in stretching the first material since conditions are the same

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A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

8 0
3 years ago
Happy Valentines Day<br><br> FR33 points for you guys
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6 0
3 years ago
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In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at
Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0
3 years ago
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