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3 years ago

A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration?

Physics

1 answer:

mr_godi [17]3 years ago

3 0

Answer:

3.33 m/s^2

Explanation:

$a = \frac{v \: - \: u}{t}$

V = final velocity ( 10m/s^2 )

U = Initial velocity ( 0m/s^2 )

t = Time taken ( 3 s )

$a = \frac{10 - 0}{3} = 3.3333......$

approximately = 3.33 m/s^2

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An electromagnet produces a magnetic eld of 0.95 T in a cylindrical region of radius 3.0 cm between its poles. A straight wire c

Daniel [21]

Answer:

0.855 N

Explanation:

Using

F = BILsinФ.................... Equation 1

Where F = magnitude of the force exerted on the wire, B = magnetic Field, I = current, L = Length of the wire in the magnetic Field, Ф = angle between the wire and the magnetic field.

Given: B = 0.95 T, I = 15.0 A, L = 2r, r = radius = 3.0 cm = 0.03 m L = 2×0.03 = 0.06 m, Ф = 90° (perpendicular)

Substitute into equation 1

F = 0.95(15)(0.06)sin90°

F = 0.855 N.

Hence the magnitude of force that is exerted on the wire = 0.855 N

6 0

4 years ago

8. Before leaving the ground an airplane traveling with constant acceleration makes a run on the

Effectus [21]

Answer:

$\color{Blue}\huge\boxed{Answer}$

<h3>B. Speed at which it leaves the ground</h3>

8 0

3 years ago

Read 2 more answers

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2

Svetradugi [14.3K]

Answer:

B. to the right

Explanation:

Given:

test charge +q

distance of the test charge from +Q, r

distance of test charge from +2Q, 2r

<u>Force on the test charge due to +Q:</u>

$F_1=k.\frac{Q.q}{r^2}$

<u>Force on the test charge due to +Q:</u>

$F_2=k.\frac{2Q.q}{(2r)^2}$

$F_2=k.\frac{Q.q}{2r^2}$

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

3 0

3 years ago

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A girl of weight 600N sits on a 3-leg chair which weighs 60N. Each chair leg presses the floor in a circle with a diameter of 5c

Daniel [21]

Pressure = total force/total area

Total force = 660 Newton's

Total area:

Each leg contacts the floor with an area of πr^2=π(0.025m)^2=0.002m^2.

Total contact area for all 3 legs = 0.006 m^2.

Pressure = (660N) / (0.006 m^2)

= 110,000 N/m^2 = 110,000 Pascal's.

8 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

Read 2 more answers