An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and

The formula for the kinetic energy of any object in motion is

                           (1/2) (mass) (velocity²).

For the object you've mentioned, it translates to

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
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Answer:

130 m/s (to two significant figures)

Explanation:

In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.

u represents the initial projectile velocity = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

A projectile's motion can be viewed as made up of independent vertical and horizontal elements.

The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.

Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.

Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.

At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s

I will analyze each option. My assumption is that the answer is C.

Option A states that gravity acts downward on the box but does not affect its horizontal acceleration, provided there is no friction.

Option B indicates that the normal force goes upward on the box, which also does not influence horizontal acceleration.

In option C, the reaction force discussed relates to Newton’s 3rd law. This reaction force acts on Lien rather than the box itself, meaning she must overcome this force to set the box in motion. I believe this is the correct choice.

Option D refers to the push force applied by her; she wouldn’t have to counteract her own force regarding the box, but must address the reaction force as I mentioned in option C.

Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m

Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
=> 152000 + 0.5 * (1.625)² * 1000 = P1 + 0.5 * (6.5)² * 1000 + (1000 * 9.81 * 1.35)
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31 - 34368.5 = 118951.81 Pa = 118.95 kPa

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = ............................1

v =

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken =

which gives us:

time taken =

so, time taken = 19.92 × seconds = 20μs.

Thus, the required time is 20 μs.