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sasho [114]
2 years ago
9

Joe pushes down the length of the handle of a 10.9 kg lawn spreader. The handle makes an angle of 45.3 ◦ with the horizontal. Jo

e wishes to accelerate the spreader from rest to 1.35 m/s in 1.6 s. What force must Joe apply to the handle? Answer in units of N
Physics
1 answer:
yuradex [85]2 years ago
7 0

Answer:

F = 13.08 N

Explanation:

given,

mass of spreader = 10.9 Kg

angle made with angle = 45.3°

accelerate the spreader to 1.35 m/s in 1.6 s.

Acceleration = \dfrac{v}{t}

                      = \dfrac{1.35}{1.6}

                      = 0.844 m/s²

momentum in = momentum out

F × t = m  × v

F cos \theta \times t = m \times v

F cos \ 45.3^0 \times 1.6 = 10.9 \times 1.35

F×1.125 = 14.175

F = \dfrac{14.175}{1.125}

F = 13.08 N

hence, the force applied by the joe is equal to F = 13.08 N

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abruzzese [7]

Answer:

The answer to your question is:   W = 390 J

Explanation:

Work is the transfer of energy when a body is moved from one place to another.

Data

Force = 65 N

mass = 45 kg

distance = 6 meters

work = ? J

Formula

W = F x d

Process

                   W = 65 N x 6 m

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2 years ago
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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
Blizzard [7]
<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M  is the mass of the asteroid

R  is the radius of the asteroid

On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

\rho=\frac{M}{V}  (2)

Finding M:

M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})  (3)

M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

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Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

Finally:

V_{esc}=117.626m/s This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

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A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
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Answer:

6.78 X 10³ N/C

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Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

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Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

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Answer:

(b) 10 Wb

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Magnetic flux is given as;

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Option "b"

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