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2 years ago

10

A 29 foot ladder leans against a building so that the angle between the ground and the ladder is 75 ∘ . How high does the ladder

reach up the side of the building? Round to 2 decimal places.

1 answer:

Aloiza [94]2 years ago

3 0

Answer:

28.01m

Explanation:

Opp/Hyp = Sin

Sin 75 = x/29

x = 29 sin 75

x = 28.01m

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A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r

NNADVOKAT [17]

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

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Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude = 2.20 mm

b) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{743}{2\pi}$

f = 118.25 Hz

c) wavelength

$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

$\lambda= 0.895\ m$

d) speed

$v = \dfrac{\omega}{k}$

$v = \dfrac{743}{7.02}$

v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

$T = \dfrac{v^2\ m}{L}$

$T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}$

T = 27.87 N

g) Power transmitted by the wave

$P = \dfrac{1}{2}m\ v \omega^2\ A^2$

$P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2$

P = 0.438 W

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2 years ago

Which of the following graphs best represents the relationship between the gravitational potential energy of a freely falling ob

VLD [36.1K]

Answer:

The last graph.

Explanation:

Gravitational potential energy is the energy possessed by a body at a given height from the Earth's surface.

The formula to find the gravitational potential energy is given as:

$U=mgh$

Where, 'U' is the gravitational potential energy.

'm' is the mass of the body.

'g' is the acceleration of the body due to gravity.

'h' is the height of the body above the Earth's surface.

So, from the above equation, it is clear that, gravitational potential energy is directly proportional to the height. So, as height increases, the gravitational potential energy increases. At the surface of Earth, where, height is 0, the gravitational potential energy is also zero.

Therefore, the correct graph is a straight line with positive slope and passing through the origin. So, the last option is the correct one.

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Answer:

1- 31.25

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3-?

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