Can someone please help me with this.

Physics

1 answer:

denis-greek [22]3 years ago

8 0

Answer:

14 fast 15 fast , not moving then constant

Explanation:

14 the slope is higher

You might be interested in

What are the two distinct ways in which energy moves outward from the solar core to photosphere?

butalik [34]

Answer:

The energy may be carried in the form of (1) radiation, where energy travels in the form of light, and (2) convection, where energy is carried by physical motion of upwelling solar gas.

Explanation:

6 0

3 years ago

The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t

frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

$Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%$

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

4 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$