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eduard
3 years ago
11

A hockey puck has a mass of 0.107 kg and is at rest. A hockey player makes a shot, exerting a constant force of 28.0 N on the pu

ck for 0.16 s. With what speed does it head toward the goal
Physics
1 answer:
gregori [183]3 years ago
5 0

Answer:

The speed does it head toward the goal = 41.87 \frac{m}{s}

Explanation:

Mass = 0.107 kg

Initial velocity ( u ) = 0

Force (F) = 28 N

Time = 0.16 sec

From newton's second law,  Force  = mass × acceleration

⇒  F = m × a

⇒  28 = 0.107 × a

⇒  a = 261.7 \frac{m}{s^{2} } --------- (1)

This is the value of acceleration.

Final speed of the mass is calculated by the equation V = U + at

⇒ U = 0 because mass in in rest position at start.

⇒ V = a t

Put the values of acceleration and time in above formula we get

⇒ V = 261.7 × 0.16

⇒ V = 41.87 \frac{m}{s}

Therefore the speed does it head toward the goal = 41.87 \frac{m}{s}

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Answer:

1. 59 Ω

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3. 0.625 kΩ

Explanation:

1. The total resistance in a series circuit is equal to the sum of the resistance.

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Therefore, the total resistance in the first circuit is 59 Ω.

2. The total resistance in a parallel circuit is equal to the sum of the reciprocals of the resistance.

\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{6.0} +\frac{1}{12} +\frac{1}{36}+\frac{1}{18} \\\frac{1}{R_T} = \frac{1}{3} \\R_T=3

Therefore, the total resistance in the second circuit is 3 Ω.

3. This is another parallel circuit, so we use the same equation from above:

\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{10} +\frac{1}{2} +\frac{1}{1} ...\\\frac{1}{R_T} =1.6\\R_T=\frac{1}{1.6}

Therefore, the total resistance in the third circuit is \frac{1}{1.6} kΩ, or 0.625 kΩ.

I hope this helps!

8 0
3 years ago
A variable that is described using both a number and direction is called
gizmo_the_mogwai [7]

they are called a vector.

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3 years ago
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A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

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2 years ago
The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t
frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

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Answer: Stronger

Explanation:

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3 years ago
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