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2 years ago

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During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)

the hammer falls in t sec is s = 16t2. what is the hammer's velocity when it strikes the ground?

2 answers:

Serga [27]2 years ago

7 0

T= 24.5 feet per second. That is the velocity it reaches at the end of its fall

Sergio [31]2 years ago

4 0

Answer:

The hammer's velocity when it strikes the ground is 128 ft/s.

Explanation:

Given that,

The distance (in feet) the hammer falls in t sec is given by the relation as :

$s=16t^2$

The hammer is accidentally dropped from a height of 784 ft. We need to find the hammer's velocity when it strikes the ground. We know that the velocity of an object is equal to :

$v=\dfrac{ds}{dt}\\\\v=\dfrac{d(16t^2)}{dt}\\\\v=32t$

When the hammer strikes ground, s = 256 ft

So,

$16t^2=256\\\\t^2=16\\\\t=4\ s$

So, the velocity of the hammer when it strikes the ground is given by :

$v=32t=32\times 4\\\\v=128\ ft/s$

So, the hammer's velocity when it strikes the ground is 128 ft/s. Hence, this is the required solution.

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A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

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Differentiating with respect to t ( time ),

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$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

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A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

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acceleration of rocket is given here as

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$v = 2.60\frac{t^2}{2}$

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here since its given that rocket will accelerate for t = 10 s

so here we have

$v = 1.30 * 10^2$

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kakasveta [241]

From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

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specific heat of water = 1.0 cal/g ∙ C°

Mass of glass = x g

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Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.

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Answer: yes.

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It depends on 3 factors:

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The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer.

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