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Mashcka [7]
2 years ago
12

A 22 kg sled is pushed for 5.2 m with a horizontal force of 20 N, starting from rest. ignore friction. find the final speed of t

he sled.
2.2 m/s
4.7 m/s
9.5 m/s
3.1 m/s
Physics
1 answer:
maksim [4K]2 years ago
5 0
We first calculate the friction on the sled using:
F = ma
a = 20 / 22
a = 0.91 m/s²

Now we use the formula:
2as = v² - u²
v = √(2as + u²)
v = √[2(0.91)(5.2) + 0]
v = 3.08 m/s
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Maru [420]

Answer:

Hello, the tripping of a 230-kilovolt transmission line.

Explanation:

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7 0
3 years ago
The volume of an object as a function of time is calculated by v = At3+B÷t where t is time measured in seconds and v is in cubic
madam [21]

Answer:

A = m³/s³ = [L]³/[T]³ = [L³T⁻³]

B = m³s = [L³T]

Explanation:

We have the equation:

V = At³ + B/t

where, the dimensions of each variable are as follows:

V = m³ = [L]³

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substituting these in equation, we get:

m³ = A(s)³ + B/s

for the homogeneity of the equation:

A(s)³ = m³

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Also,

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3 0
3 years ago
Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t
Nataliya [291]

Answer:

 Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

                                     I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

                                     Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

                                     Q = cos^-1 ( sqrt (0.464))

                                     Q = cos^-1 (0.6811754546)

                                     Q = 47.06 degrees

                                   

8 0
3 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
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Answer:

-9.46 N

Explanation:

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This value will be the projection of the force vector, on the x-axis.

For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₁ₓ, and r = F₁

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º-55º = 125º

⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N

We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₂ₓ, and r = F₂

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º + 53.3º = 233.3º

⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N

The total component of both forces along the x axis, can be found just adding both components, as follows:

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5 0
3 years ago
How many sig figs are in 3252.6
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