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2 years ago

On that system is In an isolated system, the net __ zero.

Physics

1 answer:

DedPeter [7]2 years ago

7 0

Answer:

the net work is 0

Explanation:

in an isolated system, there is no energy entering or exiting the system, and since work is dependent on the change in energy (W = ΔE) , with no change in energy, there is no work done. the net work done on or by the system must be 0

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When the Moon is directly between the Sun and Earth, a _____ tide will occur along a shoreline that is facing the Moon.

Temka [501]

<span>When the Moon is directly between the Sun and Earth, a spring tide will occur along a shoreline that is facing the Moon. </span>

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3 years ago

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

Bess [88]

Answer:

Part a)

$dF = -\frac{mv^2}{r^2} dr$

Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

$T = \frac{2\pi r}{v}$

so here if radius is constant then variation in time period is given as

$dT = - \frac{2\pi r}{v^2} dv$

8 0

2 years ago

A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0

3 years ago

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

$P=mv$

Therefore

T-Joe momentum $P_J$

$P_J=65*3$

$P_J=195N$

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2 years ago

the two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different di

svetoff [14.1K]

Answer: the block at the right lands first

Explanation:

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2 years ago