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3 years ago

I throw a ball off the edge of a 15.0m cliff. If I threw it horizontally at 8.0 m/s, how much time did it take to fall?

1 answer:

3 0

Just ignore the horizontal component

if you have a vertical displacement of 15m, 0ms^1 initial velocity, end velocity is ignored, we know the acceleration due to gravity as 9.81ms^2 so we can work out the time using SUVAT

S=15
U=0
V=?
A=9.81
T=?

S=UT + 0.5 AT^2

UT=0

therefore,

S=0.5AT^2

rearrange to:

T=SQR (2S/A)

T = 1.75 seconds

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a wall switch is connected to a wall outlet. A lamp is plugged into the same outlet. You turn the switch on, but the lamp stays

jolli1 [7]

1: only half the outlet is switched and the lamp is in the other half
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2 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

2 years ago

Which of the following paths, shown below, might light follow when passing from below water into the air?

Lelu [443]

B Because I am I really really really really really….. be it is B i am latina

6 0

2 years ago

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

Simora [160]

Answer:

0.195 m

Explanation:

Speed is distance moved per unit time, expressed as s=d/t and making d the subject of the formula then d=st

Where d is distance/depth moved, s is rhe speed of waves and t is time in seconds.

Substituting s with 1300 m/s and t with 0.00015 s then the depth of metal segment will be

D=1300*0.00015=0.195 m

Therefore, the depth is equivalent to 0.195 m

4 0

2 years ago

Which of the following is calculated by dividing work by time?

Svet_ta [14]

A is the answer i think

4 0

3 years ago