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4 years ago

Please help me with my physics! Provide work please. Question #5

Physics

1 answer:

Murljashka [212]4 years ago

6 0

#4 is correct.
For #5, don't get all tangled up. Just close your eyes, breathe deeply, and murmur "F equals M A. Force equals mass times acceleration."

The force is 15N. The mass is 12kg. You really shouldn't have any trouble calculating the acceleration.

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An object that has the ability to do work has __________ energy. (4 points)

SpyIntel [72]

Answer:

c. potential

Explanation:

6 0

3 years ago

Read 2 more answers

The intensity of a certain siren is 0.060 W/m^2 from a distance of 25 meters. What is the intensity of this siren at a distance

Nady [450]

<h2>The intensity of siren is 0.015 W m⁻²</h2>

Explanation:

The intensity of sound is inversely proportional to the square of distance between source and the observer .

Thus Intensity at a distance 25 m is I₁ ∝ $\frac{1}{r_1^2}$ I

Similarly the intensity at a distance 50 m is I₂ ∝ $\frac{1}{r_2^2}$ II

Dividing II by I , we have

$\frac{I_2}{I_1}$ = $\frac{r_1^2}{r_2^2}$ = $\frac{25^2}{50^2}$ = $\frac{1}{4}$

Thus I₂ = $\frac{1}{4}$ x 0.06 = 0.015 W m⁻²

7 0

3 years ago

A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob

andre [41]

Answer:

$I = -2.1 kg.m/s$

Explanation:

Given,

mass of the object,m = 0.30 Kg

initial speed, v_i = 3 m/s

time of collision = 0.20 s

final speed, v_f = -4 m/s

Impulse = change in momentum

$I = m (v_f-v_i)$

$I = 0.30\times (-4-3)$

$I = -2.1 kg.m/s$

Hence, impulse of the object is equal to $I = -2.1 kg.m/s$

5 0

3 years ago

How is work defined in physics

Tcecarenko [31]

Explanation:

Work is the dot product of the force and displacement vectors.

W = F · d

In other words, it is the force times the parallel component of the distance.

W = F d cos θ, where θ is the angle between the force and distance.

3 0

4 years ago

Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magni

Mkey [24]

During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s

3 0

3 years ago