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3 years ago

Briana pushed a 5 kg box of dance flyers with a force of 10 N. What is the acceleration of the box?

Physics

2 answers:

maks197457 [2]3 years ago

8 0

Answer:

Explanation:

From force =ma

a=f/m

a=10/5

a=2m/s2

fredd [130]3 years ago

3 0

F=ma ,a=f/m
F=10
m=5
10/5 = 2

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Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet

Valentin [98]

Answer:

v = 8.09 m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

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Em₀ = U = m gh

final point. To go down the slope

Em_f = K = ½ m v²

The work of the friction force is

W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

N - W_y = 0

N = W_y

X axis

Wₓ - fr = ma

let's use trigonometry

sin θ = y / L

sin θ = 11/110 = 0.1

θ = sin⁻¹ 0.1

θ = 5.74º

sin 5.74 = Wₓ / W

cos 5.74 = W_y / W

Wₓ = W sin 5.74

W_y = W cos 5.74

the formula for the friction force is

fr = μ N

fr = μ W cos θ

Work is friction force is

W_fr = - μ W L cos θ

Let's use the relationship of work with energy

W + ΔU = ΔK

-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²

v² = - 2 μ g L cos 5.74 +2 (gh)

v² = 2gh - 2 μ gL cos 5.74

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v² = 215.6 -150.16

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2 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

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Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

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The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

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Therefore the intensity at five times this distance from the source is $4W/m^2$

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Elena-2011 [213]

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