Anyone know this? PLEASE HELP, I just want to get this exam finished with ):

valentinak56 [21]

Explanation:

The factors that affect gravity are as follows:

1. mass of body

2. acceleration

Keep smiling and hope u are satisfied with my answer.Have a good day :)

6 0

3 years ago

Where are the three seismographs used to find the epicenter of this earthquake located?

iogann1982 [59]

The seismographs are located in Minneapolis, Detroit, and Charleston.

3 0

3 years ago

Read 2 more answers

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

7 0

3 years ago

Please someone helppp me.. I will mark as brainliest.

Mademuasel [1]

Answer:

b) G.P.E = Mgh

300j = M x 10 m/s² x 15 m

300 j/ 10 m/s² x 15 m = M

300j/ 150 s² = M

2kg = M

c) K.E = 1/2 m v²

K.E = 1/2 (50) (50)²

K.E = 1/2 (50) (2500)

K.E= 125000/2

K.E = 625 000 J

4 0

3 years ago

A voltage amplifier needs a high input resistance and a low output resistance. Select one: True False

cupoosta [38]

Answer:

Explanation:

In electric circuit , the potential difference is always developed across the resistance .

Now is we are to amplify the voltage , that means low input voltage is converted into high voltage output .

Therefore we require low resistance at input for low voltage and high out put resistance for high output

Thus the statement given is wrong .

6 0

3 years ago