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kherson [118]
3 years ago
15

A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat

ion when her speed is 30.0 m/s
Physics
1 answer:
kirill [66]3 years ago
4 0

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

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Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is
anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car  A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as  m_{1} \ and\ m_{2}

Let the velocity of cars A and B are denoted as v_{1} \ and\ v_{2}

The momentum before collision is-

                                                  p_{i} =m_{1} v_{1} +m_{2} v_{2}

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of  linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are v'_{1} \ and\ v'_{2}

Hence the final momentum of the system is-

                                                        p_{f} = m_{1} v'_{1} +m_{2} v'_{2}

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e      

                              p_{i} =p_{f}

                               m_{1} v_{1} +m_{2}v_{2} =m_{1} v'_{1} +m_{2} v'_{2}

Hence the option A  is right.

7 0
3 years ago
Read 2 more answers
A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer
dlinn [17]
<h2>Answer:7.14ms^{-1},4.125ms^{-1}</h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let v be the total velocity.

Given that,v=8.25ms^{-1}

We call the horizontal velocity as v_{h} and the vertical velocity as v_{v}.

v_{h}=vCos\alpha

v_{v}=vSin\alpha

where \alpha is the angle between the object and horizontal.

It is given that \alpha =30^{0}

v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}

v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}

7 0
3 years ago
An electric clothes dryer draws 15 A at 220 V. If the clothes put into the dryer have a mass of 7.0 kg
xz_007 [3.2K]

Answer:

Explanation:

7.0 - 4.0 = 3.0 kg water

3.0 kg(2.26 x 10⁶ J/kg = 6.78 x 10⁶ J

220 V(15 A) = 3300 W

1 W = 1 J/s

6.78 x 10⁶ J / 3300 J/s = 2054 s

2054 s / 60 s/min = 34 min

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2 years ago
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zmey [24]
Evolution should be the answer
5 0
3 years ago
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