A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g=

Where 1 kg=1000 g

Frequency of oscillation=

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=

Where =Angular frequency

A=Amplitude

Using the formula

Therefore, the amplitude of oscillation=A=0.199

Let us denote as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is

where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is

where a2 represents the new acceleration. 
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:

and solving for a2 gives us:

Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = &pi; r^2 = 100 &pi; =314 mm^2
(i) Stress,
&sigma;
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
&epsilon;
= ratio of extension / original length
= &sigma; / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= &epsilon; * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

Answer:

The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.

Explanation:

Given;

initial temperature, t₁ = 50 °C

final temperature, t₂ = 80 °C

Temperature change, ΔT = 80 °C - 50 °C = 30 °C

Pressure for scenario one = 1 atm

Pressure for scenario two = 3 atm

The energy needed in both scenarios is expressed as;

Where;

Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.

Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.