Ask question

Ask question

All categories

3 years ago

8

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

.1 m/s. After being thrown, the object falls freely due to gravity. Neglect air resistance and calculate the distance, in meters which the object covers between times t1 = 3.32 s and t2 = 5.08 s after it is thrown.

1 answer:

Softa [21]3 years ago

3 0

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

You might be interested in

A person hits a tennis ball with a mass of 0.058 kg against a wall.

horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

Let P is the average pressure on that area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{0.607\ N}{3\ m^2}$

P = 0.202 Pa

Hence, this is the required solution.

8 0

3 years ago

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

4 0

3 years ago

How do you do this?Or what are the formulus

LuckyWell [14K]

You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.

4 0

3 years ago

The viscosities of several liquids are being compared. All the liquids are poured down a slope with equal path lengths. The liqu

Vladimir [108]

Answer:

Move slowly and reach bottom later.

Explanation:

Viscosity is termed as the thickness or consistency of any liquid or semi liquid. It is related to the internal friction of the substance.

When several liquids are poured down with equal path lengths then the liquid will high viscosity will reach the bottom latter while one with less viscosity.

The internal friction of the molecules tends to keep them together making its consistency more thick. Thus when it will slope down from a certain height it will take more time to reach down.

5 0

3 years ago

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

VARVARA [1.3K]

Answer:

(a) f= 622.79 Hz

(b) f= 578.82 Hz

Explanation:

Given Data

Frequency= 600 Hz

Distance=1.0 m

n=120 rpm

Temperature =20 degree

Before solve this problem we need to find The sound generator moves on a circular with tangential velocity

So

Speed of sound is given by

c = √(γ·R·T/M) ............in an ideal gas

where γ heat capacity ratio

R universal gas constant

T absolute temperature

M molar mass

The speed of sound at 20°C is

c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)

c= 343.24m/s

The sound moves on a circular with tangential velocity

vt = ω·r.................where ω=2·π·n

vt= 2·π·n·r

vt= 2·π · 120min⁻¹ · 1m

vt= 753.6 m/min

convert m/min to m/sec

vt= 12.56 m/s

Part A

For maximum frequency is observed

v = vt

f = f₀/(1 - vt/c )

f= 600Hz / (1 - (12.56m/s / 343.24m/s) )

f= 622.789 Hz

Part B

For minimum frequency is observed

v = -vt

f = f₀/(1 + vt/c )

f= 600Hz / (1 + (12.56m/s / 343.24m/s) )

f= 578.82 Hz

3 0

3 years ago