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Daniel [21]
2 years ago
9

‼️HELP HELP HELP‼️

Physics
1 answer:
Reil [10]2 years ago
5 0

Answer:

sr Idontknow this

Explanation:

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A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str
ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0
3 years ago
Defination of formula bar​
Gekata [30.6K]

The Formula Bar is where data or formulas you enter into a worksheet appear for the active cell. The Formula Bar can also be used to edit data or formula in the active cell. The active cell displays the results of its formula while we see the formula itself in the Formula Bar.

mark me brainliestt :))

7 0
3 years ago
Read 2 more answers
Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,
andrew-mc [135]

Energy = (power) x (time)

-- <u>For the toaster:</u>

Power = 1.4 kW  =  1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s)  =  453,600 Joules

-- <u>For the CFL bulb:</u>

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s)  =  415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.    

-- The toaster uses 9.1% more energy than the CFL bulb.

-- The CFL bulb uses 8.3% less energy than the toaster.  

7 0
3 years ago
Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward
lbvjy [14]

Answer:

Final velocity at the top of the ramp is 6.58m/s

Explanation

Check the attachment

4 0
3 years ago
What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other
Alex73 [517]

Answer:

Electric Field = E = 36.848 N/C

Explanation:

In accordance with Columb's law

E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084

= 36948.6961 x 10^-3 = 36.848 N/C

4 0
3 years ago
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