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3 years ago

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Physics

1 answer:

Anna71 [15]3 years ago

4 0

Answer:

The total resistance of the circuit = 188ohms

Explanation:

potential difference = 25volts

For resistors in series, equivalent resistance

Rt = R1 + R2 + R3

Rt = 120 + 18 + 50

Rt = 188ohms

The total resistance of the circuit = 188ohms

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What is the mass of a 7.0 kilogram bowling ball on the surface of the moon?

babunello [35]

Answer:

1.16kg is the answer. Hope this helped

Explanation:

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2 years ago

A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle

Radda [10]

Answer:

a) x=0 %T=0, b) x= A %T=100%, c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

x = A cos (wt + Ф)

where angular velocity is related to frequency and period

w = 2π f = 2π / T

we can write the equation of the oscillation

x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

% T = 0

b) for x = A

let's find the angle

cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

% T = 100%

c) the point of maximum compression x = -A

let's look for the angles

cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

% T = 50%

6 0

2 years ago

A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$

$F_{t}-m.g=m.a$

$18.8-9.8m=5.22m$

$15.02m=18.8$

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: $F_{g}=m.g$ .

The weight for the hanging block is

$F_{g}=1.25*9.8$

$F_{g}=$ 12.25 N

Comparing tension and weight:

$\frac{12.25}{18.8}$ ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0

3 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

3 years ago

A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology a

Zigmanuir [339]

Answer:

5.95 m

Explanation:

Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g

For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,

(MV^2) / r = mg

V^2/ r = centripetal acceleration which is equal to 2g

2 × 9.8 = 10.8^2 / r

r = 116.64 /19.6

r = 5.95 m

3 0

3 years ago