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3 years ago

A 950 N skydiver jumps from an altitude of 3000 m. What is the total work performed on the skydiver?

Physics

1 answer:

pentagon [3]3 years ago

6 0

Answer:

2850000J

Explanation:

Given parameters:

Weight of diver = 950N

Altitude = 3000N

Unknown:

Total work done by the sky diver = ?

Solution:

To solve this problem, we use the expression:

Work done = Weight x height

Now insert the parameters:

Work done = 950 x 3000 = 2850000J

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A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

3 years ago

A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on

nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

5 0

3 years ago

mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

3 years ago

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable

kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

period is the time required to make a complete oscillation

T=2*π√4.9/9.81

T=2*π*0.706

T=4.44s

4.44s

5 0

3 years ago

Can someone complete this assignment will give brainliest!<br> The assignment is attached

Musya8 [376]

... I can’t see the attached assignment that you put on here.

3 0

3 years ago