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4 years ago

If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling

Physics

1 answer:

ryzh [129]4 years ago

5 0

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter. After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

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A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0

3 years ago

How strong an electric field is needed to accelerate electrons in an X-ray tube from rest to one-tenth the speed of light in a d

Bond [772]

Answer:

E= 50.1*10³ N/C

Explanation:

Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:

$vf^{2} -vo^{2} = 2*a*x$

We know that v₀ = 0 (it starts from rest), that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:

$a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2$

According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.

Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:

F = q*E (1)

In this case, q=e= 1.6*10⁻19 C

But this force, can be expressed in this way, according Newton's 2nd Law:

F = m*a (2) ,

where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.

From (1) and (2), we can solve for E, as follows:

$E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C$

⇒ E = 50.1*10³ N/C

3 0

4 years ago

The density of lead is 11.3 g/cm 3. What mass of lead is required to make a 1 cm 3 fishing sinker?

Aliun [14]

Answer:

11.3 g

Explanation:

given,

density of the lead = 11.3 g/cm³

volume = 1 cm³

mass of the lead = ?

we know,

$density=\dfrac{mass}{volume}$

mass = volume x density

mass = 1 x 11.3

mass = 11.3 g

hence, the mass of the lead is equal to 11.3 g

5 0

3 years ago

Read 2 more answers

Distinguish between basic and derived units

alex41 [277]

Answer:

Base units are defined units based on specific objects or events in the physical world. Derived units are defined by combining base units.

Base units are defined by a particular process of measuring a base quantity whereas derived units are defined as algebraic combinations of base units. For example, length is a base quantity in both SI and the English system, but the meter is a base unit in the SI system only.

8 0

3 years ago

A gas has a volume of 8.21 mL and exerts a pressure of 2.9 atm. What new volume will the gas have if the pressure is changed to

labwork [276]

Answer:

5.95 ml

Explanation:

Given info

P1=2.9 atm

V1=8.21 ml

P2=4 atm

From Boyle's law we know that p1v1=p2v2 where p and v are pressure and volume respectively. This is at a constant temperature. Making v2 the subject of formula then

V2=p1v1/p2

V2= 2.9*8.21/4=5.95 ml

3 0

3 years ago