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Ira Lisetskai [31]
3 years ago
10

I need homework help

Physics
1 answer:
KIM [24]3 years ago
5 0
1) the weight of an object at Earth's surface is given by F=mg, where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 
F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N

2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 
g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 
g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2

<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N<span>
</span>
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Which type of radio waves have the highest frequency and are used in radar detection? microwaves x-rays waves infrared waves gam
goldfiish [28.3K]

Answer: Microwaves

Explanation:

RADAR stands for radio detection and ranging.  Radio waves range from 300 GHz to 30 Hz. Modern radar uses higher frequency of waves for detection. These waves are microwaves (300 MHz-300 GHz). Microwaves are types of radio waves and are used in the radar detection.

The electromagnetic wave which has highest frequency is gamma wave but it is not used for radar detection.

Hence, the correct answer is microwaves are the type pf radio waves which have the highest frequency and are used in radar detection.

5 0
3 years ago
Read 2 more answers
For how long a time must a tow truck pull with a force of 550N on a stalled 1200kg car to give it a forward velocity of 2.0m/s?
horsena [70]

Answer:

4.36 seconds

Explanation:

According to the question;

  • Force is 550 N
  • Mass of the car is 1200 kg
  • Velocity of the car is 2.0 m/s

We are needed to find the time the car must the tow track pull the car.

  • From Newton's second law of motion;
  • Impulsive force, F = Mv÷t , where m is the mass, v is the velocity and t is the time.

Rearranging the formula;

t = mv ÷ F

Thus;

Time = (1200 kg × 2.0 m/s²) ÷ 550 N

        = 4.36 seconds

Thus, the time needed to pull the car is 4.36 seconds

5 0
3 years ago
With the water trap being used, all the possible sources of error would be eliminated.
Phoenix [80]

Answer:

True

Explanation:

7 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
2 years ago
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
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