The urethra extends from the:

P2O5 is a covalent compound used to purify sugar. What is the name of this compound?

AfilCa [17]

Answer:

B) Diphosphorus pentoxide

Explanation:

8 0

3 years ago

Read 2 more answers

ANSWER ASAP PLS

posledela

Answer:

<h3>JAWAB SECEPATNYA pliss</h3><h3 /><h3>Anda memiliki rangkaian paralel 10 volt, dengan 2 resistor di atasnya. Berapakah tegangan pada</h3><h3>resistor pertama? Di seberang kedua?</h3><h3 /><h3>(saya akan menandai tercerdas tolong bantu)</h3>

Explanation:

Hukum Ohm

= tegangan

= kuat arus

= ketahanan

Kalau kamu mau mencari tegangan listrik, kamu gunakan rumus V = I.R. Kalau ternyata kamu perlu mencari kuat arus listrik, maka gunakan rumus I = V/R. Nah, kalau yang kamu cari adalah hambatan listrik, maka gunakan rumus R = V/I.

4 0

3 years ago

If an object goes through a

aivan3 [116]

Answer:

A, The same amount of gravity

Explanation:

4 0

2 years ago

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

<span> (a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span> $F=qE$
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
</span> $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
</span> $F=qE$
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span> $F=qE$
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span> $F=qvB \sin \theta$
<span>is different from zero because v is different from zero.</span>

6 0

3 years ago

A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu

lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

$F_{frk} = -\mu_{k} * F_{n} (1)$

where μk is the kinetic friction coefficient, and Fn is the normal force.

Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

$F_{n} = F_{g} = m*g (2)$

We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

$F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)$

b)

According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
In this case, this net force is the friction force which we have already found in a).
Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
We can write the expression for a as follows:

$a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)$

c)

Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

$a = \frac{-v_{o} }{t} (5)$

Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

$t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)$

d)

From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
So, we can use the following kinematic equation in order to find the displacement before coming to rest:

$v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)$

Since the puck comes to a stop, vf =0.
Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

$\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)$

e)

The total work done by the friction force on the object , can be obtained in several ways.
One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

$W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)$

f)

By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
$P_{Avg} = \frac{\Delta E}{\Delta t} (10)$
If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

$P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)$

g)

The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

$P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)$

7 0

3 years ago