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ICE Princess25 [194]
2 years ago
12

A sky diver steps from a high-flying helicopter. if there were not air resistance, how fast would she be falling at the end of a

12 second jump?
Physics
1 answer:
Rama09 [41]2 years ago
8 0

If there is no air resistance and the body, in this case the sky diver simply falls from the helicopter then, the motion can be described as a free-falling body. The equation that would allow us to determine the speed of the body at any time, t is,

<span>                                    v(t)  = vo + gt</span>

<span>where v(t) is the unknown velocity, vo is the initial velocity which is equal to zero and g is the acceleration due to gravity which is equal to 9.8 m/s2.</span>

Substituting the known values,

<span>                                    v(t) = 0 + (9.8 m/s2)(12s)</span>

<span>                                                v(t) = 117.6 m/s</span>

<span>Hence, the speed of the sky diver at the end of 12 second is approximately 117.6 m/s.  </span>

You might be interested in
If one of the satellites is at a distance of 20,000 km from you, what percent accuracy in the distance is required if we desire
Lesechka [4]
<span>It is quite straightforward to convert an uncertainty to a percent uncertainty. We can divide the amount of uncertainty by the original amount and then multiply by 100%.

(2 m / 20,000,000 m) X 100% = 0.00001%

The percent uncertainty is 0.00001%.

The percent accuracy is the 100% - percent uncertainty.
The percent accuracy = 100% - 0.00001% = 99.99999%

The percent accuracy is 99.99999%.</span>
8 0
2 years ago
an iron rod of length 100m at 10 degree Celsius is used to measure a distance of 2km on a day when the temperature is 40 degree
german

Answer:

0.68 m

Explanation:

α = dL / L1*(dT)

dL = L1(dT) * α

Initial length, L1 = 100

Chang in Length =dL

α linear expansivity ; dL = change in length ; dT = change in temperature ; L1 = initial length

α of iron rod = 1.13 * 10^-5 k

dL = 100(40 - 10) * 1.13 * 10^-5

dL = 100(30) * 1.13 * 10^-5

dL = 3000 * 1.13 * 10^-5

dL = 3390 * 10^-5

dL = 0.0339 m

Error :

Distance measured = 2km = (2 * 1000) = 2000m

[Distance measured / (initial length + change in length)] × change in length

Error = (2000 / (100 + 0.0339)) * 0.0339

Error = (2000 / 100.0339) * 0.0339

Error = 19.993222 * 0.0339

Error = 0.6777702

Error = 0.68 m

4 0
2 years ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
lions [1.4K]

Answer:

a) Probability mass function of x

x P(X=x)

0 0.0602

1 0.0908

2 0.1700

3 0.2050

4 0.1800

5 0.1550

6 0.0843

7 0.0390

8 0.0147

b) Cumulative Distribution function of X

x F(x)

0 0.0602

1 0.1510

2 0.3210

3 0.5260

4 0.7060

5 0.8610

6 0.9453

7 0.9843

8 1.0000

The cumulative distribution function gives 1.0000 as it should.

Explanation:

Probability of arriving late = 0.43

Probability of coming late = 0.57

Let's start with the probability P(X=0) that exactly 0 people arrive late, the probability P(X=1) that exactly 1 person arrives late, the probability P(X=2) that exactly 2 people arrive late, and so on up to the probability P(X=8) that 8 people arrive late.

Interpretation(s) of P(X=0)

The two singles must arrive on time, and the three couples also must. It follows that P(X=0) = (0.57)⁵ = 0.0602

Interpretation(s) of P(X=1)

Exactly 1 person, a single, must arrive late, and all the rest must arrive on time. The late single can be chosen in 2 ways. The probabiliy that (s)he arrives late is 0.43.

The probability that the other single and the three couples arrive on time is (0.57)⁴

It follows that

P(X=1) = (2)(0.43)(0.57)⁴ = 0.0908

Interpretation(s) of P(X=2)

Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is (0.43)²(0.57)³

(ii) The probability that the two singles are on time is (0.57)²

Given that the singles are on time, the late couple can be chosen in 3 ways. The probability that it is late is 0.43 and the probability the other two couples are on time is (0.57)².

So the probability of (ii) is (0.57)²(3)(0.43)(0.57)² which looks better as (3)(0.43)(0.57)⁴ It follows that

P(X=2) = (0.43)²(0.57)³ + (3)(0.43)(0.57)⁴ = 0.0342 + 0.136 = 0.1700

Interpretations of P(X=3).

Here a single must arrive late, and also a couple. The late single can be chosen in 2 ways. The probability the person is late but the other single is not is (0.43)(0.57).

The late couple can be chosen in 3 ways. The probability one couple is late and the other two couples are not is (0.43)(0.57)². Putting things together, we find that

P(X=3) = (2)(3)(0.43)²(0.57)³ = 0.2050

Interpretation(s) P(X=4)

Since we either (i) have the two singles and one couple late, or (ii) two couples late. So the calculation will break up into two cases.

(i) Two singles and one couple late

Two singles' probability of being late = (0.43)² and One couple being late can be done in 3 ways, so its probability = 3(0.43)(0.57)²

(ii) Two couples late, one couple and two singles early

This can be done in only 3 ways, and its probability is 2(0.57)³(0.43)²

P(X=4) = (3)(0.43)³(0.57)² + (3)(0.57)³(0.43)² = 0.0775 + 0.103 = 0.1800

Interpretations of P(X=5)

For 5 people to be late, it has to be two couples and 1 single person.

For couples, The two late couples can be picked in 3 ways. Probability is 3(0.43)²(0.57)

The late single person can be picked in two ways too, 2(0.43)(0.57)

P(X=5) = 2(3)(0.43)³(0.57)² = 0.1550

Interpretations of P(X=6)

For 6 people to be late, we have either (i) the three couples are late or (ii) two couples and the two singles.

(i) Three couples late with two singles on time = (0.43)³(0.57)²

(ii) Two couples and two singles late

Two couples can be selected in 3 ways, so probability = 3(0.43)²(0.57)(0.43)²

P(X=6) = (0.43)³(0.57)² + 3(0.43)⁴(0.57) = 0.0258 + 0.0585 = 0.0843

Interpretation(s) of P(X=7)

For 7 people to be late, it has to be all three couples and only one single (which can be picked in two ways)

P(X=7) = 2(0.57)(0.43)⁴ = 0.0390

Interpretations of P(X=8)

Everybody had to be late

P(X=8) = (0.43)⁵ = 0.0147

6 0
2 years ago
un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente
Sedaia [141]

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

\Delta V = V_o\gamma \Delta T

here we know that

V_o = 40 Ltr

volume expansion coefficient of the gasoline is given as

\gamma = 950 × 10^{–6}

change in temperature is given as

\Delta T = (131 - 68) \times \frac{5}{9}

\Delta T = 35 ^oC

Now we have

\Delta V = 40(950 \times 10^{-6})(35)

\Delta V = 1.33 Ltr

3 0
2 years ago
A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x
mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0
2 years ago
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