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Mnenie [13.5K]
3 years ago
10

Volume of 10 coin is 25ml what is the volume of 1 coin

Physics
1 answer:
timama [110]3 years ago
5 0

Answer: 2.5 ML

Explanation:

10 ÷ 10 = 1

25 ÷ 10 = 2.5

The volume of 1 coin is 2.5 ML

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Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long ran
makvit [3.9K]

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

7 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
Un vehicle surt, en un instant, d'un punt A cap a un altre punt B amb una velocitat constant de 36 km/h. Després d'un minut(60 s
kati45 [8]



Ask Siri or goggle has all answers



6 0
3 years ago
Ahmad was moving to the south with v= 10 km/hr ,and Mohammed was moving with half of Ahmad's speed to the North .Write the vecto
Alexxx [7]

Answer:

5 km/hr to the South

Explanation:

VM = VA/2

VM = 10km/2hr =5km/hr

8 0
3 years ago
three carts of masses 2 kg, 18 kg, and 9 kg move on a frictionless horizontal track with speeds of 10m/s, 8m/s, and 2m/s. the ca
Elan Coil [88]

Answer:

V = 6.3 m/s

Explanation:

Given:

m₁ = 2 kg

m₂ = 18 kg

m₃ = 9 kg

V₁ = 10 m/s

V₂ = 8 m/s

V₃ = 2 m/s

__________

V - ?

Let us write the momentum conservation law for an inelastic impact:

m₁·V₁ + m₂·V₂ + m₃·V₃ = (m₁ +m₂ + m₃) ·V

Cart speed after interaction:

V = ( m₁·V₁ + m₂·V₂ + m₃·V₃ ) / (m₁ +m₂ + m₃)

V = (2·10 + 18·8 + 9·2) / ( 2 + 18 + 9) = 182 / 29 ≈  6.3 m/s

4 0
1 year ago
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