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xz_007 [3.2K]
3 years ago
13

Two forces are applied to a 2.0 kg block on a frictionless horizontal surface. F1 = 8.ON is applied to the left while F2 = 3.0 N

is applied to the right. What is the
acceleration of the block?


A.) 2.5 m/s^2 to the left

B.) 1.5 m/s^2 to the right

C.) 4.0 m/s^2 to the left

D.) 2.5 m/s^2 to the right
Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

It is C 5.0 m/s 2 to the right

Explanation:

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Water moves through a constricted pipe in steady, ideal flow. At the
Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0
3 years ago
True or False: Amplitude can increase or decrease wavelength
Marat540 [252]

Answer:

A. False, frequency can increase or decrease wavelength.

For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.

4 0
3 years ago
The energy efficiency of a car is approximately 20%. If you put 10 gallons of gas into the car, how many gallons are “wasted”?
podryga [215]
First, you find what 20% of 10 gallons of gas would be. This will show how many gallons the car actually uses.
10 gallons x 20% =
10 x 0.20 =
2 gallons used

Then you subtract that number from the total 10 gallons to get how many gallons of gas would be wasted.
10 gallons - 2 gallons = 
8 gallons of gas wasted
4 0
3 years ago
200 Coulombs of charge passes through a point in a circuit for 0.6 minutes. what is the magnitude of the current flowing​
Tasya [4]

Answer:

5.56 A

Explanation:

From the question,

Q = it.............. Equation 1

Where Q = charges, i = current, t = time.

Make i the subject of the equation

i = Q/t.............. Equation 2

Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds

Substitite these values into equation 2

i = 200/(0.6×60)

i = 5.56 A

Hence the magnitude of the current flowing through the circuit is 5.56 A

5 0
3 years ago
What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?
arlik [135]

Answer: 0.25 m/s

Explanation:  Speed = wavelengt · frequency  

v = λf   and frequency is 1/period  f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0
2 years ago
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