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3 years ago

13

Bond formed when one or more electrons are transferred from one atom to another. True or False

1 answer:

Finger [1]3 years ago

8 0

Answer: true

Explanation:

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Sarah drives her car with a constant speed of 48 miles per hour. How far can she travel in 60 minutes?

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She can dive 58 miles per hour

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3 years ago

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irina [24]

Answer:

Valley-river Landslide-Gravity Frost wedging- Glacier Canyon-Ice.

Explanation:

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3 years ago

Copper has a specific heat of 387 J/kg°C. A copper bullet is moving at 950 m/s. If the bullet is suddenly stopped so that all of

Alla [95]

Answer:

1196.02 °C

Explanation:

If the kinetic energy is converted into heat,

then,

Kinetic energy of the copper = heat energy of the copper

1/2m(v²) = cm(t₂-t₁)

where m = mass of copper, v = velocity of copper, c = specific heat capacity of copper, t₂ = final temperature of copper, t₁ = initial temperature of copper.

Since the mass of copper remains the same,

1/2v² = c(t₂-t₁)

make t₂ the subject of the equation

t₂ = 1/2(v²/c)+t₁..................... Equation 1

Given: v = 950 m/s, c = 387 J/kg°C, t₁ 30 °C

Substitute into equation 1

t₂ = 1/2(950²/387)+30

t₂ = 1196.02 °C

Hence the temperature the bullet reach before it was stopped = 1196.02 °C

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4 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

In a bicycle race on a straight road, the leader is 72.0 m from the finish line and continues to travel to the finish line with

dem82 [27]

Try this solution, answers are marked with red colour.

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4 years ago