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bearhunter [10]
3 years ago
15

Based on its location on the periodic table, an element that is not naturally occurring is?​

Physics
1 answer:
Scorpion4ik [409]3 years ago
7 0

this is a link to a web sight with a diagram to help you

https://www.google.com/url?sa=i&source=images&cd=&ved=2ahUKEwjw4v7knNDgAhVyUN8KHWgWD-wQjRx6BAgBEAU&url=%2Furl%3Fsa%3Di%26source%3Dimages%26cd%3D%26ved%3D%26url%3Dhttps%253A%252F%252Ffuturism.com%252Fwhere-do-all-the-elements-come-from%26psig%3DAOvVaw19_FOCuWs_nMsyY1YT0Da-%26ust%3D1550955269844922&psig=AOvVaw19_FOCuWs_nMsyY1YT0Da-&ust=1550955269844922

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What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?
vodomira [7]
Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2
3 0
3 years ago
Problem page a cyclist traveled 20 kilometers per hour faster than an in-line skater. in the time it took the cyclist to travel
RideAnS [48]
<span>Answer: skater x km/h cyclist 20 faster x + 20 km/h skater 30 km cyclist 80 km skater time = cyclist time t=d/r 30 / x = 80 /( x + 20 cross multiply 30 ( x + 20 ) = 80 x 30 x + 600 = 80 x 30 x - 80 x = -600 -50 x = -600 / -50 x = 12 km/h 12 km/h skater</span>
3 0
3 years ago
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of
kvv77 [185]

The  volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

According to bernaulli's theorem theorem

P+1/2pV²+pgy = constant

where p fluid density

g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.

As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is  A1V1=A2V2 .

rest is attached in image

to learn more about Pressure click here brainly.com/question/12971272

#SPJ4

4 0
2 years ago
Study the scenario.
Otrada [13]

If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.

7 0
3 years ago
Read 2 more answers
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