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Juli2301 [7.4K]
3 years ago
10

Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin

g when determining the work done on the object?
Only the component of the force perpendicular to the motion is used to calculate the work.
If the force acts in the same direction as the motion, then no work is done.
When there is an angle between the two directions, the cosine of the angle must be considered.
A force at a right angle to the motion requires the use of the sine of the angle.
Physics
2 answers:
Zigmanuir [339]3 years ago
7 0

Answer:

C

Explanation:

erastova [34]3 years ago
4 0

Answer

(C).

When there is an angle between the two directions, the cosine of the angle must be considered.


Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is W=F\times d.

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the sin(90)=0 meaning that there is no work done in the direction perpendicular to the motion.

The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is F\times d\times \cos(\theta).




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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
2 years ago
What is the value of sin square 60​
maksim [4K]

Answer:

sin {}^{2} 60 =  \frac{3}{4}

Explanation:

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5 0
3 years ago
A 45-gram object that has a volume of 9 mL has a density of
yanalaym [24]

Answer:

<h2>  The answer is </h2><h2>a. 5g/mL</h2>

Explanation:

Given data

mass m= 45g

volume v= 9mL

we know that density=m/v

substituting our given data we have

density=\frac{45}{9} =5g/mL

What is Density?

The Density of a body can be defined as the ratio of mass to volume,

or

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7 0
3 years ago
Which is the appropriate unit of measurement for pitch? decibel mel joule watt
Travka [436]
The answer would be mel
4 0
3 years ago
Read 2 more answers
A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1
baherus [9]

Answer:

The  value of the power is   P_c  =  38.55 \  W

Explanation:

From the question we are told that

   The  power  rating P_{1000} =P_b=  52 \  W

    The frequency is  f = 1000 \  Hz

    The  frequency at which the sound intensity decreases  f_k  =  20 \  Hz

     The decrease in intensity is by \beta  =  1.3 dB

Generally the  initial intensity of the speaker  is mathematically represented as

     \beta_1 =  10 log_{10} [\frac{P_b}{P_a} ]

Generally the intensity of the speaker after it has been decreased is

       \beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]

So

\beta_1-\beta_2 =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]

=>  \beta =  10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3

=>  \beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3

=>  \beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> 10log_{10} [\frac{P_b}{P_c} ] = 1.3

=> log_{10} [\frac{P_b}{P_c} ] = 0.13

taking atilog of both sides

[\frac{P_b}{P_c} ] = 10^{0.13}      

=>[\frac{52}{P_c} ] = 10^{0.13}      

=>  P_c  =  \frac{52}{1.34896}

=>   P_c  =  38.55 \  W

   

3 0
3 years ago
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