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FrozenT [24]
3 years ago
6

What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?

Physics
1 answer:
anygoal [31]3 years ago
8 0
D. 35n forwards....................
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A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h
BartSMP [9]
Specific Gravity of the fluid = 1.25 
Height h = 28 in
 Atmospheric Pressure = 12.7 psia
 Density of water = 62.4 lbm/ft^3 at 32F
 Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
 Difference in pressure = 1.26 psia
 (a) Pressure in the arm that is at Higher 
 P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
 (b) Pressure in the tank that is at Lower
 P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
4 0
3 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
3 years ago
Suppose that two objects attract each other with a gravitational force of 50N. If the mass
vagabundo [1.1K]

Answer:

112.5 N

Explanation:

50 = GMm/r^2

Let F be the new force of attraction

F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

7 0
3 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 390 N while the ot
romanna [79]

Answer:

The crate's coefficient of kinetic friction on the floor is 0.23.

Explanation:

Given that,

Mass of the crate, m = 300 kg

One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.

The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

F_1+F_2=\mu N\\\\\mu=\dfrac{F_1+F_2}{mg}\\\\\mu=\dfrac{390+320}{300\times 10}\\\\\mu=0.23

So, the crate's coefficient of kinetic friction on the floor is 0.23.

6 0
3 years ago
The attraction between two oppositely charged Atoms or groups of Atoms is which type of bond
Step2247 [10]

Answer:

The answer is ionic bond

Explanation:

8 0
3 years ago
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