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3 years ago

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An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

s at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark. What was the total distance the ant traveled?

1 answer:

11Alexandr11 [23.1K]3 years ago

4 0

Given :

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west.

It starts at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark.

To Find :

The total distance the ant traveled.

Solution :

Total distance travelled by ant = (distance between 14 and 20 inch mark) +

(distance between 20 and 16 inch mark)

Total distance = (20-14 ) + ( 20-16) = 6 + 4 = 10 inch.

Therefore, total distance the ant traveled is 10 inch.

Hence, this is the required solution.

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DESCRIBE floods and droughts. What are their effects? EXPLAIN how they can be prevented

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Answer:

<u>EFFECTS: ( doughts)</u>

There would be less water in the river for you and other people who live along the river to use.
If we use too much water during times of normal rainfall, we might not have enough water when a drought happens.

<u>EFFECTS:( FLOODS)</u>

The immediate impacts of flooding include loss of human life, damage to property, destruction of crops, loss of livestock, and deterioration of health conditions owing to waterborne diseases.
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How to prevent from flooding is :

Construct buildings above flood levels.
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3 years ago

A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di

slava [35]

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ = 14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ = $\frac{n_1sin\theta_1}{n_2}$ = $\frac{1.003*0.251}{1.33}$ = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = $\frac{4}{x_2}$ or x₂ = $\frac{4}{tan\theta_2 }$ = $\frac{4}{tan(10.86)}$ = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

Again, using the quadratic ecuation:

$t_2=24.49s$

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

$t_T=t_1+t_2=34.49s$ .

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3 years ago

A 500 kg object is hanging from a spring attached to the ceiling. If the spring constant in the spring is 900 N/kg, how far does

My name is Ann [436]

We are given that a 500 kg object is hanging from a spring. To determine the amount the spring is stretched we will use Hook's law, which states the following:

$F=kx$

Where:

$\begin{gathered} F=\text{ force} \\ k=\text{ spring constant} \\ x=\text{ distance stretched} \end{gathered}$

Since the object is hanging the only force acting on the spring is the weight of the object. The weight of the object is:

$F_g=mg$

Where:

$\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}$

Plugging in the values we get:

$F_g=(500\operatorname{kg})(9.8\frac{m}{s^2})$

Solving the operations:

$F_g=4900N$

Now we solve for "x" from Hook's law by dividing both sides by "k":

$\frac{F}{k}=x$

Now we plug in the known values:

$\frac{4900N}{900\frac{N}{m}}=x$

Solving the operations:

$5.4m=x$

Therefore, the spring is stretched by 5.4 meters.

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1 year ago

A student is standing at a distance of 45 m from a wall. He gives a loud clap at the echo is heard 0.3a later. Calculate the spe

kodGreya [7K]

Answer:

300 m/s

Explanation:

2d = vt

v = 2d/t

v = 2×90/.3

v=300 m/s

d = distance

t = time

v = velocity/speed of sound

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2 years ago