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exis [7]
3 years ago
11

Skateboard A Force: 33 N Time: 0. 12 s Skateboard B Force: 33 N Time: 0. 57 s Skateboard C Force: 33 N Time: 0. 78 s Which skate

board has the greatest impulse?.
Physics
1 answer:
Snowcat [4.5K]3 years ago
7 0

The skateboard that has the greatest impulse is Skateboard C (Impulse = 25.74 Ns)

From the question,

We are to determine which skateboard has the greatest impulse. To do this, we will calculate the impulse of each skate board.

Impulse can be calculated by using the formula

I = Ft

Where I is the impulse

F is the force

and t is the time

  • For Skateboard A

Force = 33 N

Time = 0. 12 s

∴ Impulse = 33 × 0.12

Impulse  = 3.96 Ns

  • For Skateboard B

Force = 33 N

Time = 0. 57 s

∴ Impulse = 33 × 0.57

Impulse = 18.81 Ns

  • For Skateboard C

Force = 33 N

Time = 0. 78 s

∴ Impulse = 33 × 0.78

Impulse = 25.74 Ns

Hence, the skateboard that has the greatest impulse is Skateboard C (Impulse = 25.74 Ns)

Learn more here: brainly.com/question/21840495

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crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

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7 0
4 years ago
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

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Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

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Length of rod  = length of pendulum - radius of bob

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8 0
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Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

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K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

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When you drop a 0.43 kg apple, Earth exerts
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Answer:

The acceleration of the earth is 7.05 * 10^-25 m/s²

Explanation:

<u>Step 1:</u> Data given

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mass of earth = 5.98 * 10 ^24 kg

<u>Step 2:</u> Calculate the acceleration of the earth

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a(earth) = F(apple/earth)/m(earth)

a(earth) = 4.214N /5.98 * 10 ^24 kg

a(earth) = 7.05 * 10^-25 m/s²

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