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natta225 [31]
2 years ago
5

In which direction is there a net force of 200 N?

Physics
2 answers:
hram777 [196]2 years ago
4 0

Answer:

right    

Explanation:

The up and down forces cancel

  then you are left with a    -200    and a  + 400

     resulting in a + 200   (to the right)

sweet [91]2 years ago
3 0

Answer:

It's Left

Explanation:

have a nice day!! :D

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6 latter word and it has a s and a I and it has mass (9.______________liquids, and gases all have mass.)
Firdavs [7]

Answer:

Solids

Explanation:

Solids, liquids, and gases all have mass.

7 0
2 years ago
A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
BartSMP [9]

Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

5 0
3 years ago
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that
Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

T = \frac{t}{n}

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

T = \frac{135\ s}{98}

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

T = 2\pi \sqrt{\frac{l}{g}}

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}

<u>g = 11.2 m/s²</u>

3 0
3 years ago
How could you record the number 4000 and report 2 significant figures?
yawa3891 [41]

Explanation:

Write in scientific notation.

4000 = 4.0×10³

5 0
3 years ago
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