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3 years ago

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Albert views Henry moving by at a constant velocity. Henry has a light clock and also a stopwatch which were synchronized when a

t rest. As Albert views these two moving clocks, which of these statements would be true according to Albert? The light clock runs slow compared to the stopwatch. The stopwatch runs slow compared to the light clock. The two are still synchronized but run fast. The two are still synchronized but run slow.

1 answer:

slavikrds [6]3 years ago

3 0

The two are still synchronized, but both are running slow.

(So is Henry's heartbeat, breathing rate, and all the frequencies in the hip-hop .mp3 he's listening to.)

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If your weight is 100N and you run up a flight of stairs that is 6 m high and it takes

fenix001 [56]

Answer:

power=300watt

Explanation:

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2 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

3 years ago

The product of charge through, and potential across, an electrical device is:

melomori [17]

The option is Work.

The product of charge and potential is equal to the energy. Adn, as we know work is related to energy as the capacity to do work.

Alos, because, Potential is given as, V = E/q
or E = Vq
Thus, t<span>he product of charge through, and potential across, an electrical device is:work
</span>

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2 years ago

Read 2 more answers

Suppose A=BnCm, where A has dimensions LT, B has dimensions L2T-1, and C has dimensions LT2. Then the exponents n and m have the

julsineya [31]

Explanation:

The expression is :

$A=B^nC^m$

A =[LT], B=[L²T⁻¹], C=[LT²]

Using dimensional of A, B and C in above formula. So,

$A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}$

Comparing the powers both sides,

2n+m=1 ...(1)

2m-n=1 ...(2)

Now, solving equation (1) and (2) we get :

$n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}$

Hence, the correct option is (E).

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2 years ago

Using Ohm’s Law, explain how resistance changes in relation to current, assuming that voltage remains constant.

Cerrena [4.2K]

V=IR so voltage is directly proportional to current. So for a given resistance increasing the voltage will result in a high current as well. This is because resistance is proportional to the voltage over the current. Ex: I=V/R
Hope this helped. THANKYOU for asking. <span />

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2 years ago