Ask question

Ask question

All categories

3 years ago

11

As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens th

at Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 39.0 m/s.
How fast is the watermelon going when it passes Superman?

1 answer:

SSSSS [86.1K]3 years ago

3 0

Answer:

The velocity of watermelon when it passes Superman is 78 m/s.

Explanation:

Height of the building, d = 320 m

Speed of the superman, v = 39 m/s

We need to find the speed of watermelon when it passes Superman. Let t is the time taken by the watermelon. So,

$d=ut+\dfrac{1}{2}gt^2$

Here u = 0

$d=\dfrac{1}{2}gt^2$

$39t=\dfrac{1}{2}\times 9.8\times t^2$

t = 7.95 seconds

Let v is the speed of the watermelon. It is given by :

$v=gt$

$v=9.8\times 7.95$

v = 77.91 m/s

or

v = 78 m/s

So, the velocity of watermelon when it passes Superman is 78 m/s.

You might be interested in

The gravitional energy of an object is always measured realative to the ?

zaharov [31]

I would say that the answer would be MASS.

4 0

3 years ago

Read 2 more answers

I NEED HELP ASAP

LiRa [457]

The object is called a meteor because it is producing Streak of light and has not yet struck earth.

<h3><u>Explanation:</u></h3>

A meteoroid is a celestial object which is very smaller than an asteroid. These objects are produced as a collision impact from mars or moon and float freely in space without any specific orbit. When they come inside the Earth's gravitational field, they are attracted by the Earth's gravity to Earth's crust. These objects in Earth's atmosphere are called meteors. As they travel through Earth's atmosphere, they do face a huge friction from Earth's atmosphere which let them burn and that is visible as the tail of the meteor.

Most of them are so small that they are burnt away in the atmosphere. But some are bigger and they reach the Earth's surface and are called as meteorites.

4 0

3 years ago

Read 2 more answers

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

2 years ago

an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o

ZanzabumX [31]

Answer:

The maximum amplitude ( $V_{max}$ ) will be 7.96 V.

Explanation:

We know, for distortion free operation, the slew rate (S) of an OPAMP is written as

$S = 2 \pi f V_{max}$

where ' $f$ ' is the highest frequency signal.

Therefore, from the above equation we can write,

$&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V$

3 0

3 years ago

Debido al desorden en el laboratorio un científico tiene 2 termómetros diferentes pero no sabe en qué escalas están por lo que d

just olya [345]

Answer:

La escala del termómetro ''A'' es grados Celsius.

La escala del termómetro ''B'' es grados Fahrenheit.

Explanation:

Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.

Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.

Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :

$F=(\frac{9}{5})C+32$ (I)

Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos $C=18$ en la ecuación (I) deberíamos obtener $F=64.4$ ⇒

$F=(\frac{9}{5}).(18)+32=32.4+32=64.4$

Efectivamente obtenemos el valor esperado. Finalmente, corroboramos que la temperatura del termómetro ''A'' está medida en grados Celsius y la temperatura del termómetro ''B'' en grados Fahrenheit.

6 0

3 years ago