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lakkis [162]
3 years ago
11

A sledder is at rest atop a 5.0-m high hill. She’s texting and doesn’t realize her sled starting to slip, moving her to the bott

om, and then up to the top of the adjacent 3.0-m high hill. How fast is the she going at point (B)? Ignore friction. (round to the nearest hundreth; use g=10m/s^2).

Physics
1 answer:
djyliett [7]3 years ago
5 0

Answer: V = 7.75m/s

Explanation:

In this problem, we will consider conservation of energy where gravitational potential energy is converted into kinetic energy.

Please find the attached file for the solution.

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Which of the following contains the majority of the mass in the solar system?
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First think which has less mass in the solar system. The sun is the largest object in the solar system, so the answer is C. the sun. Hope I helped! :P
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A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
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To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

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3 years ago
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal sp
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Answer:

C

Explanation:

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only vertical speed changes

so H speed will stay 30 m/s

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An object is most likely to sink in water if
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A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under
Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0
4 years ago
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