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3 years ago

Please answer this soon. Thank u.

Physics

1 answer:

vodomira [7]3 years ago

6 0

Answer:

The distance used to calculate the pressure of atmosphere is 75 cm.

Explanation:

Mercury barometer is used to find the pressure of the atmosphere where it is placed.
As shown in the figure, the barometer has an inverted tube where mercury has risen by 75 cm. This height is taken into count to measure the atmospheric pressure.
At the place with high pressure the mercury in the inverted tube will go very high and vice versa.
Hence to find the pressure we must note down the height of mercury that has risen the inverted tube.

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In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of

seraphim [82]

Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

8 0

2 years ago

Calculate the total displacement of a mouse walking along a ruler, if it begins at the x=5cm, and then does the following: It wa

Lana71 [14]

<span>To begin, the mouse walks from 5 to 12 cm, for a displacement of 7 cm. Next, it walks 8 cm in the opposite direction, for a total displacement of (7 + [-8]) or (-1) cm. This leaves the mouse on 4 cm, and then it walks from there to the 7cm location, for a displacement of 7-4 or +3 cm. Adding 3cm to -1cm gives a final displacement of +2cm.</span>

6 0

3 years ago

4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling

pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

plug in known varibles,

v=2.4*8

v=19.2m/s

4 0

2 years ago

Question 2 of 10

Mekhanik [1.2K]

the awnser to ur question is D

6 0

3 years ago

Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

6 0

3 years ago