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Sidana [21]
3 years ago
10

Here is the picture of an aircraft, with the forces working on it shown. Identify the force that corresponds to the question mar

k.
A. Buoyancy
B. Atmospheric pressure
C. Gravity
D. Density

Physics
2 answers:
mel-nik [20]3 years ago
8 0
Your answer is option c
Alisiya [41]3 years ago
3 0
Answer- c: gravity
explanation- it just usb
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What is the force on a person’s hand, which is using a rope to accelerate a 5 kg block upward with an acceleration of 2.2 m/s 2
slavikrds [6]

Answer:

F = 11 N

Explanation:

Given,

Mass of a block, m = 5 kg

Acceleration of the block, a = 2.2 m/s²

We need to find the force on the person's hand. Let it is F. We know that force is given by the product of mass and acceleration as follows :

F = ma

F = 5 kg × 2.2 m/s²

F = 11 N

So, the force on a person's hand is 11 N.

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3 years ago
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Answer:

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Explanation:

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Why does the evaporation of sweat cool down your skin?
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3 years ago
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A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an
mina [271]

Answer:

μ = 0.33

Equal to 3.2 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

Normal force N pushing up.

Weight force mg pulling down.

Friction force Nμ pushing opposite the direction of motion.

Sum of forces in the y direction.

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction.

∑F = ma

Nμ = ma

Substitute.

mgμ = ma

μ = a/g

μ = (3.2 m/s²) / (9.8 m/s²)

μ = 0.33

As found earlier, the acceleration is a = gμ.  Since g and μ are constant, a is also constant, so it does not change with velocity.

5 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area= \sigma = 1.99*10^{-7} C/m^2

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates isE = \frac{\sigma}{\epsilon}

force acting on charge is F = q E

Work done by charge q id\Delta X =\frac{ q\sigma \Delta x}{\epsilon}

this work done is converted into kinectic enerrgy

\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}

solving for v

v = \sqrt{\frac{2q\Delta x}{\epsilon m}

\epsilon = 8.85*10^{-12} Nm2/C2

v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}

v = 1.15*10^{7} m/s

8 0
3 years ago
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