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3 years ago

It is impossible to destroy __________.

Physics

2 answers:

fredd [130]3 years ago

6 0

C.energy
The law of conservation of energy says that energy can neither be created nor destroyed.

beks73 [17]3 years ago

3 0

The applicable term that fills up the problem is C. energy. This is derived for the law of conservation or the first law of thermodynamics which states that energy cannot be created nor destroyed just transformed to another state. This notion is the basis of all transformations and laws existing

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What effect does a catalyst have on a system in equilibrium?

Jet001 [13]

ANSWER:

What effect does a catalyst have on a system in equilibrium?

The system is unaffected.

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3 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

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3 years ago

What type of acceleration does an object moving in a circular path with constant speed experience?

Marysya12 [62]

An object moving in a circular path has centripetal acceleration. <em>(A)</em>

7 0

3 years ago

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Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

mash [69]

Answer:

The value is $} N = 66 \ blocks$

Explanation:

From the question we are told that

The weight of the block is $W_b = 100 \ N$

The dimension of the block is $d = 0.400 m \ X \ 0.250 \ m \ X \ 0.130 \ m$

Generally two atmosphere is equivalent to

$P_{2atm} = 2 * 1.013 *10^{5} = 202600 \ N/m^2$

Generally 1 atm = $1.013 *10^{5} N/m^2$

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

$A = 0.250 * 0.130$

=> $A = 0.0325 \ m^2$

Generally the force due to this blocks is mathematically represented as

$F = N * W_b$

Here N is the number of blocks

$} 202600 = \frac{N * 100 }{ 0.0325}$

=> $} N = 66 \ blocks$

3 0

3 years ago

A 0.290 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato

Sergeu [11.5K]

Answer:

A) The speed of the potato at the lowest point of its motion is 7.004 m/s

B) The tension on the string at this point is 8.5347 N

Explanation:

Here we have that the height from which the potato is allowed to swing is 2.5 m

Therefore we have ω₂² = ω₁² + 2α(θ₂ - θ₁)

Where:

ω₂ = Final angular velocity

ω₁ = Initial angular velocity = 0 rad/s

α = Angular acceleration

θ₂ = Final angle position

θ₁ = Initial angle position

However, we have potential energy of the potato

= Mass m×Gravity g× Height h

= 0.29×9.81×2.5 = 7.1125 J

At he bottom of the swing, the potential energy will convert to kinetic energy as follows

K.E. = P.E. = 7.1125 J

1/2·m·v² = 7.1125 J

Therefore,

v² = 7.1125 J/(1/2×m) = 7.1125 J/(1/2×0.290) = 49.05

∴ v = √49.05 = 7.004 m/s

B) Here we have the tension given by

Tension T in the string = weight of potato + Radial force of motion

Weight of potato = mass of potato × gravity

Radial force of motion of potato = mass of potato × α,

where α = Angular acceleration = v²/r and r = length of the string

∴ Tension T in the string = m×g + m×v²/r = 0.290×(9.81 + 7.004²/2.5)

T = 8.5347 N

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3 years ago

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