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3 years ago

Objects falling through air experience a type Of friction called ____

Physics

2 answers:

DaniilM [7]3 years ago

6 0

Answer: That's air resistance.

Explanation: Well, air resistance is an upward force exerted on falling objects.

( I hope this helped <3 )

77julia77 [94]3 years ago

4 0

Answer:air resistance

Explanation:

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blagie [28]

Answer:

Hold on Ill answer it..When do u need it by

Explanation:

4 0

2 years ago

B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

$V_{int}=\sqrt{U^2+19.6H}$

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration is mathematically given as

$a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2$

Where

acceleration is still -9.81 m/s2,

Hence,

$V^2-U^2=2(-9.81)*-H$

Therefore

$V_{int}=\sqrt{U^2+19.6H}$

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7 0

2 years ago

When an object radiates heat, the strength of this radiation far from the object decreases when distance from the source increas

Andrew [12]

Answer:

The source of cosmic background radiation filled the entire universe.

Explanation:

D:The source of cosmic background radiation filled the entire universe.

7 0

2 years ago

Read 2 more answers

A proton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the pro

svlad2 [7]

Answer:

Magnitude of the force on proton = F = 1.1085 × 10^-15 N

Explanation:

Charge on proton = q = 1.60 × 10^-19 C

Velocity of proton = V = 4.0 × 10^4 m/s

Magnetic field = B = 0.20 T

Angle between V and B = θ = 60

We know that,

F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)

F = 1.1085 × 10^-15 N

8 0

3 years ago

Read 2 more answers

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

2 years ago