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serious [3.7K]
3 years ago
12

Is velocity a scalar or a vector

Physics
2 answers:
Zielflug [23.3K]3 years ago
7 0
A vector because is going in a direction 
gizmo_the_mogwai [7]3 years ago
5 0
Velocity<span> is a</span>vector<span> quantity; it is direction-aware.</span>
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In a cell, the amount nutrition coming in equals the amount of waste going out. This is an example of _____.
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5 0
3 years ago
A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?
rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

8 0
3 years ago
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You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha
UNO [17]
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6 0
3 years ago
Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200
andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

                                   ΔK.E = ΔP.E

                                  K_2 - K_1 = P_1- P_2

Where,  K_2: Kinetic energy of hammer head as it hits the I-beam

             K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

             P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

             P_1: Initial gravitational potential energy of hammer head      

- The expression simplifies to:

                                K_2 = P_1

Where,                     0.5*m*v2^2 = m*g*s12

                                v2 = √(2*g*s12) = √(2*9.81*3)

                                v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

                               Wnet = ( P_3 - P_1 ) + W_friction

                               Wnet = m*g*s13 + F*s23

                               n*s23 = m*g*s13 + F*s23

Where,    n: average force the hammerhead exerts on the I-beam.

               s13 = s12 + s23

Hence,

                             n = m*g*( s12/s23 + 1) + F

                             n = 200*9.81*(3/0.074 + 1) + 60

                             n = 81562 N

                               

                                                   

6 0
3 years ago
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