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?

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:

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

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