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4 years ago

13

Which location is MOST LIKELY to be struck by a hurricane? A) Charleston, South Carolina B) San Diego, California C) Buffalo, Ne

w York D) Wichita, Kansas

1 answer:

xxTIMURxx [149]4 years ago

6 0

It is A) Charleston because it is the closest city to warm Atlantic waters.

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Help me with this question please

Salsk061 [2.6K]

Answer:

5.en

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Explanation:

<h3>#CARRY ON LEARNING</h3><h3>#BRAINLITS </h3>

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3 years ago

A 1000kg ar accelerates from rest to 25.0m/s in 4.20

bezimeni [28]

Answer:

74.4 kilowatts or 99.8 horsepower

Explanation:

The explanation is in the attachment.

7 0

3 years ago

Two people person A and person B lift the same amount of weight but person A lifts it in a shorter amount of time who has more p

const2013 [10]

Person A has more muscular strength then person B. B has muscular endurance.

3 0

3 years ago

When we look into the sky every day we get to see the results of all the behaviors of waves. The blue color of the sky results f

mr_godi [17]

Answer:

λ = 0.4 x 10⁻⁶ m = 400 nm

Explanation:

The relationship between frequency, wavelength and speed of an electromagnetic wave is given as follows:

$c = f\lambda$

where,

c = speed of light = 3 x 10⁸ m/s

f = frequency of the light wave = 7.5 x 10¹⁴ Hz

λ = wavelength of the light = ?

Therefore,

$3\ x\ 10^8\ m/s = (7.5\ x\ 10^{14}\ Hz)\lambda\\\\\lambda = \frac{3\ x\ 10^8\ m/s}{7.5\ x\ 10^{14}\ Hz}$

<u>λ = 0.4 x 10⁻⁶ m = 400 nm</u>

4 0

3 years ago

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

$1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s$

After this time, we expect a horizontal displacement of 12 meters, so that $v_0$ satisfies

$v_0(0.57\,\mathrm s)=12\,\mathrm m$

$\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}$

5 0

4 years ago