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3 years ago

6

Consider a simple harmonic oscillator made from a block of mass m attached to a ng constant k. What changes could you make that

would double its maximum speed. By how much would this change increase its energy?

1 answer:

faust18 [17]3 years ago

8 0

Answer:

energy would become four times

Explanation:

Kinetic energy K= 0.5mv^2

⇒ $v=\sqrt{\frac{2m}{K} }$

now, v'= 2v

$2v=v'=2\sqrt{\frac{2m}{K} }$

v'= $\sqrt{\frac{2\times4E}{m} }$

hence, speed can be doubled by four folding the energy or by decreasing mass to one fourth.

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The total pressure for a fluid is a. the sum of the hydrostatic and static pressures. b. the sum of the hydrostatic and dynamic

m_a_m_a [10]

<h2>The option a is most appropriate </h2>

Explanation:

The total pressure due to liquid column at any place is the sum of

( i ) pressure due to liquid column called hydrostatic pressure

( ii ) the pressure due to air column above the liquid column , which is called the static pressure

Thus total pressure is the sum of hydrostatic and static pressure .

Thus the option a is most appropriate

3 0

3 years ago

What is the equivalent resistance of a circuit that contains two 50.00

Whitepunk [10]

Answer:

A

Explanation:

Resistors in series add. There is only one path the current can take. That's why Christmas Tree lights sometimes give a lot of trouble. If a bulb burns out, it could be any one of them and time is needed to find the burned out bulb.

That being the case R = R1 + R2

R1 = 50 ohms

R2 = 50 ohms

R = 50 + 50

R = 100 ohms

Answer A

4 0

4 years ago

Read 2 more answers

pav-90 [236]

Answer:

Total energy is constant

Explanation:

The laws of thermodynamics state that thermal energy (heat) is always transferred from a hot body (higher temperature) to a cold body (lower temperature).

This is because in a hot body, the molecules on average have more kinetic energy (they move faster), so by colliding with the molecules of the cold body, they transfer part of their energy to them. So, the temperature of the hot body decreases, while the temperature of the cold body increases.

This process ends when the two bodies reach the same temperature: we talk about thermal equilibrium.

In this problem therefore, this means that the thermal energy is transferred from the hot water to the cold water.

However, the law of conservation of energy states that the total energy of an isolated system is constant: therefore here, if we consider the hot water + cold water as an isolated system (no exchange of energy with the surroundings), this means that their total energy remains constant.

4 0

3 years ago

A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s^2. The friction experienced by the swimmer is 100 N. How many N

Helga [31]

Answer:

490N

Explanation:

According Newton's second law!

\sum Force = mass × acceleration

Fm - Ff = ma

Fm is the moving force

Ff s the frictional force = 100N

mass = 65kg

acceleration = 6m/s²

Required

Moving force Fm

Substitute the given force into thr expression and get Fm

Fm -100 = 65(6)

Fm -100 = 390

Fm = 390+100

Fm = 490N

Hence the force that will cause two cart to move is 490N

5 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago