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2 years ago

How do objects and substances in motion have kinetic energy

Physics

1 answer:

Fiesta28 [93]2 years ago

6 0

Answer:

Explanation:

Kinetic energy is one of the types of mechanical energy and it is the energy due to the object in motion.

Kinetic energy is given as 1/2mv^2

A body cannot have a K.E when it is not in motion because of the velocity

Velocity is the change is position of an object or particle with time taken, so if the object doesn't change it position then it won't have velocity.

kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

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En una barra de 6m que se utiliza como palanca se coloca el fulcro a 2 m de distancia del extremo derecho, como se muestra en la

Anna007 [38]

Answer:

Fp = 45 N

VMR = 2

VMI = 2

Efficiency = 100%

Explanation:

English Translation

In a 6m bar that is used as a lever, the fulcrum is placed 2m away from the right end, as shown in the figure. At that same end it is required to support a load of 90N. Neglecting lever weight, determined Fp, VMI, VMR and system efficiency

Solution

With the diagram not massively needed to solve this question.

Although, the load of 90 N is placed at the right end of the bar, the fulcrum is placed 2 m from that right end and the balancing force or effort is placed at the extreme most left end of the 6 m bar.

Taking moment about the fulcrum,

The sum of clockwise moments must balance the sum of anti-clockwise moments.

(Load) × (Distance of load from the fulcrum) = (Balancing force) × (Distance of balancing force from fulcrum)

Load = L = 90 N

Distance of load from the fulcrum = 2 m

Balancing force or Effort = Fp = ?

Distance of balancing force from fulcrum = 6 - 2 = 4 m

90 × 2 = Fp × 4

Fp = (180/4) = 45 N

The mechanical advantages are them given as

VMR = (Load)/(Effort) = (90/45) = 2

VMI = (Distance of Effort from the fulcrum)/(Distance of Load from fulcrum) = (4/2) = 2

Efficiency is then given as the VMR divided by VMI in percentage terms

Efficiency = 100% × (VMR/VMI)

Efficiency = 100% × (2/2) = 100%

Hope this Helps!!!

4 0

3 years ago

Your friend wants to be magician and intends to use Earth’s magnetic field to suspend a current-carrying wire above the stage. H

sesenic [268]

Answer:

$I=1960A$

Explanation:

Under this condition, . In order to suspend the wire, this magnetic force would have to be equal in magnitude to the gravitational force exerted by Earth on the wire the maximum force at angle

F=ILxB

Now to the suspend the wire so use the maximum force and solve to the current knowing the magnetic field of the earth

∑Fy=0

$F_{m}-F_{g}=0$

$I*L*\beta-m*g=0$

Solve to I current

$I=\frac{m*g}{L*\beta}$

$I=\frac{10x10^{-3}kg*9.8m/s^2}{1m*0.5x10{-4}T}$

$I=1960A$

I suggest do an toher act is really risk that current for an act

7 0

3 years ago

A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost

densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0 (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = $\frac{1}{2}$ m₁u₁² + $\frac{1}{2}$ m₂u₂² ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = ( $\frac{1}{2}$ x 2.0 x 5.0²) + ( $\frac{1}{2}$ x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = $\frac{1}{2}$ (m₁ + m₂)v² ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = $\frac{1}{2}$ ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

3 0

3 years ago

What type of material is found in the asthenosphere?

IceJOKER [234]

Answer:

The correct answer is option C.

Explanation:

The layer below the lithosphere , the upper layer of the mantle of the earth with high heat pressure .It is made up of almost solid rocks but also flows not which indicates that it consist of viscoelastic fluid of molten rocks.This is because in some regions there can be presence of molten rocks.

6 0

3 years ago

Read 2 more answers

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c

Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

$\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}$

hence, the speed of the spheres is 37.45 m/s

8 0

3 years ago