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2 years ago

A person on a merry-go-round is constantly changing direction. True or false

Physics

2 answers:

Elden [556K]2 years ago

6 0

True. Because you’re always facing a different direction

guapka [62]2 years ago

6 0

True, you are spinning around

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the difference between any two successive numbers always seems to be _ more than the preceding difference

erma4kov [3.2K]

Here's li $^{}$ nk to the answer:

cutt $^{}$ .ly/4Rq $^{}$ tIvk

6 0

2 years ago

Read 2 more answers

What is the current in a 160V circuit if the resistance is 2Ω? V= I= R =

sveta [45]

Answer:

80 amperes

Explanation:

Current in the circuit = ?

Voltage in the circuit = 160 Volts

Resistance = 2 Ω

Voltage = Current x Resistance

V = IR

160V = I x 2 Ω

I = 160V / 2 Ω

I = 80 Amperes

Therefore the current in the circuit is 80 amperes :)

8 0

3 years ago

1. A rock of granite has a mass of 50 kg. if it’s weight in water

igor_vitrenko [27]

Answer:

The first part of the question is asking about BUOYANT FORCE or UPTHRUST.

Upthrust =TRUE WEIGHT-APPARENT WEIGHT

TRUE WEIGHT=mg

TRUE weight=50kg×10m/s²

=500N

upthrust=500N-380N

FB=120N

volume of the rock=mass/density.

since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.

upthrust=Vdg

120N=V×1000kg/m³×10m/s²

120N=V×10000kg/m²s²

120/10000=V

v=0.012m³

please mark brainliest, hope it helped

6 0

2 years ago

Interpretar el servicio de la verdad

Lelechka [254]

Umm what are you trying to say

6 0

2 years ago

(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?

malfutka [58]

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

$T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}$

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

$\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T$

Thus, the new period will be reduced by 50%

8 0

3 years ago