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malfutka [58]
3 years ago
8

Different between pressure and force​

Physics
1 answer:
9966 [12]3 years ago
3 0

Force is mass into acceleration

and pressure is force applied per unit area.

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A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied
faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

3 0
3 years ago
A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor
natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = 120\Omega

Fixed resistance, R = 120\Omega

Gauge Factor, G.F = 2.01

Supply Voltage, V_{s} = 3\ V

Strain, \epsilon = 1000\times 10^{-6}\ strain

Now,

To calculate the output voltage, V_{o}:

WE know that strain is given by:

\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}

Thus

V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}

Now, substituting the suitable values in the above eqn:

V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}

V_{o} = 1.507\ mV

6 0
3 years ago
Which of the following has to happen before the eardrum begins to vibrate with the same frequency as the source of the sound wav
777dan777 [17]
I just took the test it is D




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8 0
3 years ago
Read 2 more answers
Red clothes will---the red light.A)Reflect B)refract C)Absorb D)transmission E)Dispersion
posledela

Red clothes look red  because they REFLECT the red light, and absorb light of other colors.

7 0
3 years ago
Physics double pivot question​
andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

3 0
3 years ago
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