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3 years ago

Help me please,

Physics

1 answer:

garik1379 [7]3 years ago

7 0

Option B

Explanation:

no distance was given only the acceleration due to the fact that it went up (10m/s/s)

s0 it is

0 m/s and 10m/s/s (option B)

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Where danger is and how to avoid it as well as to locate other whales

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If a honeybee is flying 7.0 m/s, what is the kinetic energy (Joules) if its mass is 0.1 g?

svetlana [45]

Answer:

KINETIC ENERGY=K.E=0.00245 Joules ≈ 0.003 Joules

Explanation:

mass=m=0.1g=0.0001kg

velocity=v=7.0m/s

kinetic energy=K.E.=?

as we know that

$kinetic energy=\frac{1}{2}mv^2\\ putting values\\kinetic energy=\frac{1}{2}(0.0001kg)(7.0m/s)^2\\ kinetic energy=\frac{1}{2} 0.0049joules\\kinetic energy= 0.00245Newton meter or joules$

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3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

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3 years ago

What happens at<br> the Bernoulli effect

Paul [167]

Answer:

Within a horizontal flow fluid, points of higher fluid speed will have less pressure than points of slower fluid speed.

Explanation:

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3 years ago

you weight 650 N. What would you wieght if the Earth were four times as massive as it is and its raduis were three times its pre

LiRa [457]

To solve this problem we will apply the concept given by the law of gravitational attraction, which properly defines gravity under the function

$g = \frac{GM}{r^2}$

Here,

G = Gravitational Universal Constant

M = Mass of Earth

r = Distance between the human and the center of mass of the Earth

The acceleration due to gravity when is 4 times the mass of Earth and 3 times the radius would be given as,

$g_1 = \frac{GM}{r^2}$

$g_2 = \frac{G(4M)}{(3r)^2}$

$g_2 = \frac{4GM}{9r^2}$

$g_2 = \frac{4}{9} g_1$

The weight is defined as

$W = mg_1$

So the new weight would be given as

$W' = mg_2$

$W' = m(\frac{4}{9} g_1 )$

$W' = \frac{4}{9} mg_1$

$W' = \frac{4}{9} W$

$W' = \frac{4}{9}(650)$

$W' = 288.8N$

Therefore the weight under this condition is 288.8N

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3 years ago