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siniylev [52]
3 years ago
13

"the velocity of the block is constant, so the net force acting on the block must be zero. Thus the normal force equals the weig

ht and the force of friction equals the applied force" what, if anything, is wrong with this statement
Physics
1 answer:
RSB [31]3 years ago
6 0

Answer:

Explanation:

"the velocity of the block is constant, so the net force acting on the block must be zero. Thus the normal force equals the weight and the force of friction equals the applied force" The only thing which is wrongly stated here is that the normal force equals the weight. Normal force is not always equal to weight. If external force is applied at some angle with the horizon , normal force may be more or less than weight. Even in case of an object placed on an  inclined plane , weight is not equal to normal force .

All other parts of the statement are correct.

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Help please! Will give brainliest to first correct answer!
Olegator [25]

Answer:

1) Addition of a catalyst

2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)

Explanation:

1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.

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8 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

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4 years ago
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Sample Response: "No, steel and carbon would not form metallic bonds because metallic bonds only form between metals. Iron is a metal, but carbon is not."

7 0
3 years ago
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Answer:

the answer to the question is the part that is reflected by clouds and atmosphere 26%

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