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2 years ago

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The steel plate is 0.3 m thick and has a density of 7850 kg>m3 . determine the location of its center of mass. also find the

reactions at the pin and roller support.

1 answer:

cestrela7 [59]2 years ago

3 0

answer with full explanation is attached below

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List some reasons why growth characteristics are more useful on agar plates than on agar slants

SpyIntel [72]

Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates.

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2 years ago

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A kitten sits in a lightweight basket near the edge of a table. A person accidentally knocks the basket off the table. As the ki

Lesechka [4]

Answer:

the answer is B

Explanation: this was actually an ap exam question a few years back. the reason for answer B is that the only force being applied to the kitten is the force of gravity after being pushed.

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2 years ago

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2

Svetradugi [14.3K]

Answer:

B. to the right

Explanation:

Given:

test charge +q

distance of the test charge from +Q, r

distance of test charge from +2Q, 2r

<u>Force on the test charge due to +Q:</u>

$F_1=k.\frac{Q.q}{r^2}$

<u>Force on the test charge due to +Q:</u>

$F_2=k.\frac{2Q.q}{(2r)^2}$

$F_2=k.\frac{Q.q}{2r^2}$

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

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2 years ago

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In a later chapter we will be able to show, under certain assumptions, that the velocity v(t) of a falling raindrop at time t is

lianna [129]

Answer:

$\lim_{t \to \infty} v(t) =vT$

Explanation:

Using distributive propierty:

$v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}$

So:

$\lim_{t\to \infty} vT-e^{-gt}$

The limit of the sum of two functions is equal to the sum of their limits, therefore:

$\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}$

The limit of a constant function is the constant, hence:

$\lim_{t\to \infty} vT=vT$

Now, let's solve the other limit:

$\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}$

The limit of a constant times a function is equal to the product of the constant and the limit of the function, so:

$\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)$

$-g(\infty)=-\infty$

Therefore:

$e^{(-\infty)} =0$

Finally:

$\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT$

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2 years ago

A man attempts to pick up his suitcase of weight w_s by pulling straight up on the handle. (Part A figure) However, he is unable

Stels [109]

Answer:

B. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

Explanation:

As we know that man tried to pull the suitcase in upward direction

so here we can say that net upward force on the suitcase applied by the man is not as sufficient that it will counter balance the weight of the suitcase

So here the net force on the suitcase will be zero and the force equation is given as

$F + F_n = W$

so we have

$F_n = W - F$

so correct answer will be

B. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

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2 years ago