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3 years ago

12

Which of the following is maintained across the terminals of a battery? A. a potential difference B. a voltage drop C. an electr

ic charge D. both A and B

2 answers:

Simora [160]3 years ago

8 0

both a and b

ur answer is d

serious [3.7K]3 years ago

5 0

I believe the answer is C. An electric charge

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Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at

inysia [295]

Answer:Impossible to answer without knowing their actual initial speeds

Explanation:

Given

Ball A launched at angle of $20^{\circ}$

Ball B is launched at angle of $75^{\circ}$

At the highest point Ball vertical velocity will be and there will be only horizontal velocity i.e. $v_acos20$ and $v_bcos75$

where $v_a and v_b$ is the velocity of A & B

thus to find which one is greater we need to know their initial velocity.

It is impossible to know which one have higher velocity at highest point.

7 0

3 years ago

Calculate the value of δs when 64.0 g of ga(l) solidifies at 29.8 ∘c.

frez [133]

The variation of entropy of a substance is given by
$\delta S = \frac{\delta Q}{T}$ (1)
where
$\delta Q$ is the heat exchanged in the process
T is the absolute temperature at which the transformation occurs.

The process in the problem is the solidification of the liquid Gallium, which releases an amount of heat equal to:
$\delta Q = m L_f$
where m is the mass of the substance and $L_f = 80.1 J/g$ is the latent heat of fusion of Gallium. Using m=64.0 g, we find
$\delta Q= m L_f = (64.0 g)(80.1 J/g)=-5126.4 J$
where the negative sign means the Gallium is releasing heat to the environment.

Now we can use equation (1) to find the variation of entropy, but first we need to convert the temperature into Kelvin:
$T=29.8^{\circ} + 273.15 = 302.95 K$

And so the variation of entropy is
$\delta S = \frac{\delta Q}{T}= \frac{-5126.4 J}{302.95 K}=-16.92 J$
and the negative sign means the entropy in the process is decreasing.

7 0

3 years ago

Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce

Stells [14]

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

$\omega=\frac{v}{r}$

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

$|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}$

the maximum relative velocity is 0.04m/s

hope this helps!!

6 0

3 years ago

Read 2 more answers

A gymnast with mass m1 = 44 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =

SVEN [57.7K]

Answer:

Explanation:

e. At the right edge of the beam

Check attachment for other solution

4 0

3 years ago

A body moves due north with velocity 40 m/s. A force is applied

marshall27 [118]

Answer:

the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Explanation:

The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:

$\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v$

p: momentum

m: mass

$\Delta v$ : change in the velocity

The sign of the change in the velocity determines the direction of rate of change. Then you have:

$\Delta v=v_2-v_1$

v2: final velocity = 35m/s

v1: initial velocity = 40m/s

$\Delta v =35m/s-40m/s=-5m/s$

Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

4 0

3 years ago