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MariettaO [177]
3 years ago
9

What must two objects have for gravity to exist between them

Physics
2 answers:
jolli1 [7]3 years ago
4 0
Mass and Distance. I hope I helped! :)
Musya8 [376]3 years ago
3 0
The two factors are mass and distance between them.
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When an electron moves from a lower to a higher energy level the electron?
vladimir1956 [14]
A photon is absorbed

your question is incomplete, please consider revising it.
8 0
3 years ago
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How much brighter is a Sun-like star than the reflected light from a planet orbiting around it?
zloy xaker [14]

Answer:

E) a billion times brighter

Explanation:

  • <u>The sun is a star, which is about billion times brighter as the reflected light from any planet orbiting around it. </u>
  • The brightness is based on its composition and its position from the planet. The sun happens to be the brightest star on the Earth's sky which is about 13 billion times brighter than the next brightest star.
8 0
3 years ago
A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing
Damm [24]

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, \tau _{shaft} = 30 MPa

Shearing stress of key, \tau _{key} = 240 MPa

width of key, w = \frac{d}{4}

d is shaft diameter

Solution:

Torque, T = \frac{P}{\omega }

where,

\omega = \frac{2\pi  N}{60}

T = \frac{7460}{\frac{2\pi  (1200 )}{60}} = 59.365 N-m

Now,

\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}

30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}

d = 0.0216 m

Now,

w =  \frac{d}{4} =  \frac{0.02116}{4} = 5.4 mm

Now, for shear stress in key

\tau _{key} = \frac{F}{wl}

we know that

T = F \times r =  F. \frac{d}{2}

⇒ \tau _{key} = \frac{\frac{T}{\frac{d}{2}}}{wl}

⇒ 240\times 10^{6} = \frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}

length of the rectangular key, l = 4.078 mm

7 0
3 years ago
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Suppose you studied the velocity of a cheetah overtime after it starts running to chase an deer. After the linear fit, a spreads
EastWind [94]

Explanation:

While studying the velocity of a cheetah over time in a spreadsheet program is given by :

y = 2.2 x + 1.2 ...(1)

We know that,

v=v₀+at ...(2)

v₀ is velocity when t = 0, v is velocity after time t, a is acceleration and t is time.

If we consider time t on x-axis and v on y axis, then only we can draw a plot of equation (2). On comparing equation (1) and (2) we get :

a = 2.2 (but it is not correct as we don't know about axes).

Hence, the correct option is (a) "We cant tell without knowing what was plotted on the horizontal and vertical axes"

6 0
4 years ago
A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev
Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf   where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²  

a= 4×π²×2×3²

a=710.6 m/s²

5 0
3 years ago
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