Formula of atmospheric pressure

You are in the lab and are given two rods set up so that the top rod is directly above the bottom one. The two straight rods 50-

kompoz [17]

Answer:

Explanation:

The magnetic force due to lower rod must be equal to weight of upper rod for equilibrium .

magnetic field due to lower rod on upper rod

= ( μ₀ / 4π ) x(2i / r ) , i is current , r is distance between rod

= 10⁻⁷ x 2 x 15 / 1.5 x 10⁻³

= 20 x 10⁻⁴ T

force on the upper rod

= B i L , B is magnetic field , i is current in second rod and L is its length

= 20 x 10⁻⁴ x 15 x .50

= 150 x 10⁻⁴ N

= .015 N

This force can balance a wire having weight equal to .015 N .

= .00153 kg

= 1.53 g .

wire should weigh 1.53 g .

3 0

4 years ago

A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock

lbvjy [14]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

$V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s$

Part B. Using Kinematics expression for distance, time and initial velocity

$y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s$

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

8 0

3 years ago

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging

bekas [8.4K]

Answer:

The answers to the question are

(a) 2.1 m/s

(b) 0.83 N

(c) 1.9 N

Explanation:

To solve the question, we list out the varibles

Length, l of string = 0.8 m

mass of rock, m = 0.12 kg

Angle with the verrticakl, θ = 45 °

a) To find the speed of the rock when the string passes through the vertical position we have

From the first law of thermodynamics

Potential energy = kinetic energy

m×g×l×(1-cosθ) = 1/2×m×v²

That is v² = 2×g×l×(1-cosθ)

= 2×9.81×0.8×(1-cos45) = 4.597

or v = √4.597 = 2.1 m/s

(b) The tension in the string when it makes an angle of 45∘ with the vertical is given by

For balance between Tension and mass of rock is gigen by

∑Forces = 0, T - m×g×cosθ = 0

or T = m×g×cosθ = 0.12×9.81×cos45 = 0.83 N

c) The tension in the string as it passes through the vertical

when passing through the vertical we have T - m×g = (m×v²)/r

or T = m×g + (m×v²)/r = mg(1+2(1-cosθ)) =0.981*0.12 (1+ 2(1-cos45)) =1.867 N

= 1.9 N

3 0

3 years ago

At cruise conditions, air flows into a jet engine at a steady rate of 60 lbm/s. Fuel enters the engine at a steady rate of 0.59

Amanda [17]

Answer:

ρ=0.0102lbm/ft^3

Explanation:

To solve this problem we must take into account the equation of continuity, this indicates that the sum of the mass flows that enter a system is equal to the sum of all those that leave.

Therefore, to find the mass flow of exhaust gases we must add the mass flows of air and fuel.

m=0.59+60=60.59lbm/s( mass flow of exhaust gases)

The equation that defines the mass flow (amount of mass that passes through a pipe per unit of time) is as follows

m=ρVA

Where

ρ=density

V=velocity

m=mass flow

A=cross-sectional area

solving for density

ρ=m/VA

ρ=60.59/{(1485)(4)}

ρ=0.0102lbm/ft^3

8 0

3 years ago

A racing car’s velocity is increased from 44 m/s to 66 m/s in 11 seconds. What is the acceleration? and what is the displacement

Musya8 [376]

Answer:

Explanation:

Acceleration is defined as the change in velocity per unit of time

Acceleration (a) = ΔV/t

V = Velocity t = time

ΔV = V₂ - V₁ t = 11s

V₁ = 44m/s

V₂ = 66m/s

ΔV = 66 - 44

= 22m/s

Acceleration (a) = 22/11

= 2m/s²

Displacement (d): Displacement equals the original velocity multiplied by time plus one half the acceleration multiplied by the square of time.

d = v₀ + 1/2*at²

v₀ = 44m/s

a = 2m/s

t = 11s

d = 44 + 1/2*(2 x 11²)

= 44 + 121

d = 165m

8 0

4 years ago

Formula of atmospheric pressure​

Formula of atmospheric pressure