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2 years ago

5

Calculate the distance the marble travels during the first 3.0 seconds. [Show all work, including the equation and substitution

with units.]

1 answer:

Alenkinab [10]2 years ago

3 0

D = V0t + 0.5at^2

Where d is the distance

V0 is the initial velocity

A is the acceleration

T is time

From the graph a = 4/3 m/s2

D = 0(3) + 0.5( 4/3 m/s2) ( 3 s)^2

D = 6 m

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A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0

2 years ago

A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if

saul85 [17]

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

F = 0.15×10×2×sin0

F = 0

Goodluck

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2 years ago

An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of

satela [25.4K]

<span>First, we use the kinetic energy equation to create a formula: Ka = 2Kb 1/2(ma*Va^2) = 2(1/2(mb*Vb^2)) The 1/2 of the right gets cancelled by the 2 left of the bracket so: 1/2(ma*Va^2) = mb*Vb^2 (1) By the definiton of momentum we can say: ma*Va = mb*Vb And with some algebra: Vb = (ma*Va)/mb (2) Substituting (2) into (1), we have: 1/2(ma*Va^2) = mb*((ma*Va)/mb)^2 Then: 1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2 We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1: 1/2(ma) = (ma^2)/mb Cancel the ma of the left, leaving the right one with exponent 1: 1/2 = ma/mb And finally we have that: mb/2 = ma mb = 2ma</span>

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2 years ago

a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr

mash [69]

Answer:

the inductive reactance of the coil is 1335.35 Ω

Explanation:

Given;

inductance of the coil, L = 250 mH = 0.25 H

effective current through the coil, I = 5 mA

frequency of the coil, f = 850 Hz

The inductive reactance of the coil is calculated as;

$X_l = \omega L = 2\pi f L\\\\X_l = 2\pi \times 850 \times 0.25\\\\X_l = 1335.35 \ ohms$

Therefore, the inductive reactance of the coil is 1335.35 Ω

6 0

1 year ago

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

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2 years ago