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MrRa [10]
2 years ago
8

The additional energy in a beaker of hot water compared to an otherwise identical beaker of room temperature water is _ _ _ _ _

_ __. Potential energy work thermal energy chemical energy.
Physics
2 answers:
Fynjy0 [20]2 years ago
7 0
<span>The additional energy in a beaker of hot water compared to an otherwise identical beaker of room temperature water is thermal energy.
hope this helps</span>
il63 [147K]2 years ago
6 0

Correct answer choice is :


C) Thermal energy


Explanation:


Thermal energy is a term applied loosely as an analog for more rigorously specified thermodynamic measures such as the physical energy of a system heat or tangible heat, which are described as examples of change of energy or for the specific energy of a degree of choice in a thermal system. Heat energy that is converted into light can be obtained in many ways. It can be generated by heating fuels such as coal, oil, gas or wood. It can also be obtained from steam from a geothermal field or produced by nuclear effects.

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A company designed and sells an ultrasonic​ receiver, which detects sounds unable to be heard by the human ear. The receiver can
Ksenya-84 [330]

Answer:

28√3 m

Explanation:

A = vertex where receiver is placed

S = focus

Bp = r = radius of the outside edge

Bc = 2r = diameter

The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.

7 0
2 years ago
(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang
avanturin [10]

6m at 0.9 radian means 6m at 51.57^0

Since positive angles are counter clockwise

6m at 51.57^0 can be written as 6*cos51.57^0i+6*sin51.57^0j = 3.73i+4.70j

5m at -75^0 can be written as 5*cos(-75)^0i+5*sin(-75)^0j = 1.294i-4.83j

4m at 1.2 radian means 4m at 68.75^0

4m at 68.75^0 can be written as 4*cos68.75^0i+4*sin68.75^0j = 1.45i+3.73j

6m at -210^0 can be written as 6*cos(-210)^0i+6*sin(-210)^0j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = \sqrt{1.274^2+0.6^2} = 1.98 m

c) Angle,  tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0

7 0
2 years ago
A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a
Phantasy [73]

The question is incomplete. It comes with a set of answer choices.


These are the answer choices:


Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.


Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.


Both Alex and his friend observe it as 10 m/s.


Both Alex and his friend observe it as less than 10 m/s.



Answer: Both Alex and his friend observe it as 10 m/s.


Justification:


1) The speed is relative to the frame of reference.


2) It is said that the both Alex and his friend are standing still.


3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).


Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0
2 years ago
Read 2 more answers
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