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FrozenT [24]
3 years ago
10

Barycenter can be described as??

Physics
1 answer:
lesya692 [45]3 years ago
7 0

Answer: 3) the point between 2 objects where they balance each other

Explanation:

The center of mass of two bodies about which the bodies orbit and balance each other is known as Barycenter. It is the point of balance of two or more celestial bodies. The Barycenter determines the orbit of the celestial bodies.

For example: the binary stars orbit a common center of mass. Pluto and its moon Charon revolve about their Barycenter.

Hence, the correct option is 3.

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While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali
fredd [130]

The question is incomplete, the options are;

RI^2

I^2/R

R/I^2

R/V^2

RV^2

V^2/R

VI

VIR

Select all that apply

Answer:

P=RI^2

P=V^2/R

P=VI

Explanation:

Power is the rate at which energy is changing in a circuit. It is shown by the formulas outlined above from the group of answer choices. Since the current (I), voltage (V), and resistance (R) were mentioned in the question, any of three three formulas could be used to obtain the power drawn by the conductor.

7 0
2 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
4. Uncle Harry weighs 180 pounds. What is his mass in kilograms?
MatroZZZ [7]
180 pounds (lb) converts to 81.647 kilograms (kg).
5 0
2 years ago
Which of the following is a physical property of wood
Black_prince [1.1K]
The density, hard, strong, and rough.
3 0
3 years ago
Read 2 more answers
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