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4 years ago

CD bisects AB at point G. If AE = BE, which equation must be true?

Physics

2 answers:

ratelena [41]4 years ago

8 0

AG = BG equation must be true

rodikova [14]4 years ago

3 0

Answer: The correct answer is option D.

Explanation:

It is given that line CD is bisecting the side AB in triangle ABE

and AE = BE

Bisecting a line segment means that diving a line segment into two equal parts. So, when CD bisects side AB at point G it means that:

AG = BG

Hence, the correct answer is option D.

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you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

Tomtit [17]

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

$W=Fd$

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

$d=\frac{h}{sin \theta}$ is the distance covered, where

h = 1 m is the vertical rise

$\theta=30^{\circ}$ is the slope of the plane

Substituting and solving, we find

$W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J$

Learn more about work:

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5 0

3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circuit with two lightbulbs rated at 3.0

Blababa [14]

Answer:

$R = 24 ohm$

Explanation:

As we know that terminal voltage and EMF of cell is given as

$V = EMF - ir$

$ir = 12 - 11.8$

$ir = 0.2 Volts$

resistance of two bulbs is given as

$R = \frac{V^2}{P}$

$R = \frac{12^2}{3}$

$R = 48 ohm$

now these two bulbs are connected in parallel

so equivalent resistance is given as

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{48} + \frac{1}{48}$

so we have

$R = 24 ohm$

3 0

3 years ago

Some ideal gas is constrained in the V"part of anadiabatic container of volume V" V$. The rest of the container is vacuum. When

solmaris [256]

Answer:

Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:

∆S = NK*ln(V"V$/V").

Where V"V$ is final Volume (Vf) after constraint's removal,

V" is Initial Volume (Vi) before constraint's removal.

Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations

Explanation:

Given in the question, the container is an adiabatic container.

For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).

Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")

5 0

3 years ago

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t

diamong [38]

Answer:

11.98 N

Explanation:

Normal force = mg = 2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

.602 = F / (2.03 * 9.81) = 11.98 N

5 0

3 years ago

A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potentia

Llana [10]

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

$V = \frac{kQ}{r}$

$V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}$

V = 4.23 x 10⁷ volts

V = 42.3 MV

5 0

4 years ago