When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh

Answer:

Explanation:

The equation used to determine the maximum height of the bowling pin during its trajectory is given by;

H = u²/2g

where u, the initial speed/velocity, equals 10m/s

g stands for gravitational acceleration = 9.81m/s²

Substituting in the values gives us

H = 10²/2(9.81)

H = 100/19.62

Consequently, the highest point of the bowling pin's center of mass is approximately 5.0m.

We will utilize Wien's displacement law, given by the equation λ T = b, where λ represents the wavelength of emitted light from a heated object at maximum. By substituting the provided temperature and constant b into the equation, we find λ for various temperatures: at 500 K, λ = 5.796 μm or 5796 nm; at 1050 K, λ = 2760 nm; at 1800 K, λ = 1610 nm; and at 2500 K, λ = 1159.2 nm. The visible light spectrum starts at 740 nm, suggesting that at 2500 K, some visible red light may emerge as its calculated peak wavelength is within the visible range.

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  

Explanation:

The question indicates that

    The wire’s diameter is  

     The radius of the wire is  

     Aluminum's resistivity is

       The electric field variation is described as

         

     

The charge is effectively given by the equation

       

Where A is the area expressed as

       

 Thus,

       

Therefore

     

By substituting values

     

     

The question states that t =  5 seconds

           

         

         

     

According to the second law, heat, often called thermal energy, cannot be entirely turned into work.
The second statement is closely tied to this law.
We can conclude that some energy dissipates while some is used for work.

Answer: c. 4.56 × 105 J

Explanation:

Given the mass of the lead brick, m = 7.25 kg

Starting temperature T1 = 18.0 °C

Ending temperature T2 = 328 °C

The specific heat capacity for lead, c = 128 J/(kg∙°C)

And the latent heat of fusion Lfusion = 23,200 J/kg

The required energy Q =?

Using the following equations

Energy required, Q = mc (T2 - T1) + mLfusion

Substituting in the values we have: 7.25 kg * 128 J/(kg∙°C) * (328 - 18°C) + 7.25 kg * 23200 J/kg

= 455880 J

= 4.56 x 10^5 J