Best definition of an electromagnetic wave?

During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t

tangare [24]

Answer:

<u>77.8 m/s, downward</u>

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

Average speed = distance / time

Then:

(Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

Vf + Vi = 90.634 equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

Vf - Vi = 64.876 equation 2

Adding the two equations:

2Vf = 155.510

Vf = 77.8 m/s downward (velocities must be reported with their directions)

8 0

3 years ago

What likely happens to the hart rate as the cardiac mucle gets stronger

Kaylis [27]

The heart rate will likely decrease. As the cardiac muscle, or heart, gets stronger, it takes less effort to pump more blood. As a result, the heart will probably beat less, decreasing the heart rate. This is why athletes often have lower heart rates than the average person.

5 0

3 years ago

An elevator is moving down with an acceleration of 3.36 m/s2.

sergeinik [125]

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

$\sf\longrightarrow Weight = mg - ma \\$

$\sf\longrightarrow Weight = m ( g - a ) \\$

Mass of the man = 64.2 kg

$\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\$

$\sf\longrightarrow Weight = 64.2 * 6.44 N\\$

$\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\$

5 0

2 years ago

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago