An object of mass 12kg is held at a weight of 5m above the ground for 30s. Calculate the workdone during this period.

Two cars collide at an intersection. Car A , with a mass of 2000kg , is going from west to east, while car B , of mass 1400kg ,

ahrayia [7]

Complete Question:

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

Answer:

a) 6.36 m/s b) 4.57 m/s

Explanation:

a) Assuming no external forces acting during the collision, total momentum must be conserved.

As momentum is a vector, we can decompose it along two directions perpendicular each other.

Just for convenience, we choose as our x-axis to the W-E direction, and as our y-axis, the direction N-S.

If we know that total momentum must be conserved, same must be true for both components, px and py.

Applying the information provided (both cars become enmeshed after the collision, moving at an angle of 65º south of east from the point of impact), we have:

px = ma * va = (ma+mb) * vab * cos 65º (1)

py = mb * vb = (ma + mb) * vab * sin 65º (2)

Replacing by the values of ma, mb, and sin 65º, we can solve for vab, as follows:

vab = 1,400 kg* 14.0 m/s / (3,400 Kg * sin 65º) = 6.36 m/s

b) Replacing vab from above in (1), and solving for va, we have:

va = 3,400 kg* 6.36 m/s* cos 65º / 2,000 Kg = 4.57 m/s

7 0

3 years ago

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago

How<br> are<br> earthquakes measured quantitatively

AlexFokin [52]

Answer:Intensity: The severity of earthquake shaking is assessed using a descriptive scale – the Modified Mercalli Intensity Scale. Magnitude: Earthquake size is a quantitative measure of the size of the earthquake at its source. The Richter Magnitude Scale measures the amount of seismic energy released by an earthquake.

Explanation:

4 0

3 years ago

I need to know what a-e is. Please and thank you!

beks73 [17]

The different types of energy transfers are convection, conduction, and radiation.
I'm not too sure on the second one but thermodynamics relates thermal energy, kinetic energy, and potential energy. I'm basing this on the formulas of the laws of thermodynamics, but it could also be temperature, heat, or work.

6 0

3 years ago

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica

Evgen [1.6K]

This question is incomplete, the complete question is;

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?

(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)

Answer: the maximum charge q is 716.85 μF

Explanation:

Given data;

with = 3.0 cm = 0.03

breathe = 5.0 m

Area = 0.03 × 5 = 0.15 m²

dielectric strength E = 1.00 × 10⁸

∈₀ = 8.85 × 10⁻¹²

constant K = 5.4

maximum charge = ?

the capacitor C = KA∈₀ / d

q = cv so c = q/v

now

q/v = KA∈₀ / d

q = vKA∈₀/d = EKA∈₀

we substitute

q = (1.00 × 10⁸) × 5.4 × 0.15 × 8.85 × 10⁻¹²

q = 716.85 × 10⁻⁶ F

q = 716.85 μF

the maximum charge q is 716.85 μF

7 0

3 years ago

An object of mass 12kg is held at a weight of 5m above the ground for 30s. Calculate the workdone during this period.​

An object of mass 12kg is held at a weight of 5m above the ground for 30s. Calculate the workdone during this period.