PHYSICS COLLEGE

*Which of the following cannot be an example of projectile motion
A. A football flying through the air
B. An apple falling from a tree
C. A pencil rolling on the ground
D.A rocket dropping from its maximum height

Answers

Answer 1
Answer: A. football flying through the air
Answer 2
Answer: A cause it flying through the also a projectile is a object flying in the air like a arrow for example/also can I get Brainly

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Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!

Answers

Answer:

F_N=1234.8N

Explanation:

Hello.

In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:

F_N=63kg*9.8m/s^2+500N\n\nF_N=617.4N+500N\n\nF_N=1234.8N

Best regards.

If a charge is located at the center of a spherical volume and the electric flux through the surface of the sphere is φ o, what is the flux through the surface if the radius of the sphere doubles?

Answers

Answer:

The electric flux remains unchanged

Explanation:

From Gauss law the Electric flux is directly proportional to the number of electric field lines passing through a surface. The number of field lines passing through a surface become if the radius is doubled becomes 1/4th that is when radius of the Gaussian surface is doubled, but at the same time, the surface area has increased 4 times , so the electric flux remains unchanged

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

Plzzz helppppp!!! I need answers A, B, C & D

Answers

Answer: the answer i for c is yes 0& 10

Explanation:

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

Explanation:

The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow

putting the values

π x .2945² x 12.1

= 3.3  m³ /s

To know the pipe's diameter at the refinery we shall apply the following formula

a₁ v₁ = a₂ v₂

a₁ v₁ and  a₂ v₂ are volume rate of flow of liquid which will be constant .

3.3 = a₂ x 6.29

a₂ = .5246 m³

π x r² = .5246

r = .4087 m

= 40.87 cm

diameter

= 81.74 cm

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

Answers

Answer:

1.6* 10^(-7) N

2.4* 10^(-7) N

Explanation:

i_(1) = 1 A

i_(2) = 4 A

r = distance between the two wire = 5 m

F = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

F = (\mu _(o))/(4\pi )(2i_(1)i_(2))/(r)

F = (10^(-7))(2(1)(4))/(5)

F = 1.6* 10^(-7) N

r'} = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

B = (\mu _(o))/(4\pi ) \left \left ( (2i_(2))/(r') \right - (2i_(1))/(r') \right \right ))

B = (10^(-7)) \left \left ( (2(4))/(2.5) \right - (2(1))/(2.5) \right \right ))

B = 2.4* 10^(-7)
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