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steposvetlana [31]
4 years ago
11

You’re squeezing a springy rubber ball in your hand. If you push inward on it with a force of 1 N, it dents inward 2 mm. How far

must you dent it before it pushes outward with a force of 5 N?
Physics
1 answer:
creativ13 [48]4 years ago
8 0

Answer:

10mm

Explanation:

According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.

Mathematically, F = ke where;

F is the applied force

k is the elastic constant

e is the extension

from the formula k = F/e

k = F1/e1 = F2/e2

Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm

Let F1 = 1N e1 = 2mm

The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?

Substituting this values in the formula above we have

1/2=5/e2

Cross multiplying;

e2 = 10mm

This shows that we must have dent it by 10mm before it pushes outwards by a 5N force

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A copper rod has a length of 1.3 m and a cross-sectional area of 3.6 10-4 m2. One end of the rod is in contact with boiling wate
sdas [7]

To solve the problem it is necessary to apply the concepts related to heat flow,

The heat flux can be defined as

\frac{dQ}{dt} = H = \frac{kA\Delta T}{d}

Where,

k = Thermal conductivity

A = Area of cross-sectional area

d = Length of the rod

\Delta T= Temperature difference between the ends of the rod

k =388 W/m.\°C Thermal conductivity of copper rod

A = 3.6 *10^{-4} m Area of cross section of rod

\Delta T=100-0=100\°C Temperature difference  

d=1.3m length of rod

Replacing then,

H = \frac{kA\Delta T}{d}

H = \frac{(388)(3.6 *10^{-4})(100)}{1.3}

H=10.7446J

From the definition of heat flow we know that this is also equivalent

H = \dot{m}*L

Where,

\dot{m} = Mass per second

L = 334J/g Latent heat of fusion of ice

Re-arrange to find \dot{m},

H = \dot{m}*L

\dot{m}=\frac{L}{H}

\dot{m}=\frac{334}{10.7446}

\dot{m} = 31.08g/s

\dot{m}= 0.032g/s

Therefore the mass of ice per second that melts is 0.032g

7 0
3 years ago
Write one advantage of MKS system over CGS system.​
Temka [501]
More convenient
More commonly used
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3 years ago
PLEASE HELP WILL MAKE BRAINIEST (GIVING AWAY 20 POINTS)
Bond [772]

1) The most potential energy is at position A

2) The most kinetic energy is at position C

Explanation:

1)

The gravitational potential energy is the energy possessed by an object due to its position in a gravitational field, and it is given by the equation

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object relative to the ground

From the equation, we see that the potential energy is directly proportional to the heigth of the object: therefore, the roller coaster in this problem will have the most potential energy at its highest postion, so at position A.

2)

The total mechanical energy of the roller coaster at any point along the track is given by

E=PE+KE

where

PE is the potential energy

KE is the kinetic energy

Assuming there is no friction, the mechanical energy E is constant. This means that when PE increases, KE decreases, and when PE increases, KE decreases.

Therefore, the cart will have maximum kinetic energy when the potential energy is at minimum: and since the potential energy is directly proportional to the height of the track, this will occur at the lowest position, so at position C.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

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brainly.com/question/10770261

#LearnwithBrainly

5 0
3 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
Help with this review pls!!
o-na [289]
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5 0
3 years ago
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