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Lady_Fox [76]
2 years ago
6

José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo

ved without the riders falling. When the ride accelerates to a speed of 17.0 m/s, his angular momentum is 3570 kg m2/s. What is his mass?
Remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits
Physics
1 answer:
zaharov [31]2 years ago
8 0

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


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Water evaporates off lakes. Winds blow across the planet. Where does the energy come from for these and other weather processes?
Otrada [13]

Answer:

B. Solar energy

Explanation:

The water cycle is driven primarily by the energy from the sun. This solar energy drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.

8 0
2 years ago
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A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0
Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

F_{lift}=F_{up} - F_{dw}=0.5*p*V^2

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

F_{lift}=0.5*p*V^2=W

By replacing the known values it is possible to find the plate's weight:

F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W

W=72.6 N

When the plate kept to 30° from the vertical the moment equation balance is written as:

F_{lift}=0.5*p*V^2=W*sen(30\°)

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

V=\sqrt(W*sen(30\°)/0.5p)

V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})

V=\sqrt(60.5 \frac{N}{kg/m^3})

V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})

V=\sqrt(60.5 \frac{m^2}{s^2})

V= 7.78 m/s

7 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
What is the minimum work needed to push a 950-kg car 710 m up along a 9.0° incline? ignore friction?
kakasveta [241]
<span>1.0344645 MJ The minimum energy need is the potential energy of the car at the top of the ramp and is given by mass*gravity*height mass is known, gravity is assumed to be 9.81m/s^2 as it is on earth, and height must be calculated using trigonometry. height=sin(9 degrees)*710m=111meters so potential energy = 950kg*111m*9.81m/s^2=1.0344645 MJ Using the law of the conservation of energy we can assume that the energy expended to push the car up the incline was at least the potential energy gained by moving 111m against the pull of gravity.</span>
7 0
2 years ago
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When particles are connected by a cable, the motions of the particles are _________________. always independent random always de
White raven [17]

Answer:

Option C (always dependent) would be the appropriate answer.

Explanation:

  • The movements including its particles everywhere are depending on when particles were also associated with either a cable, since, if we want the substances to modify their location, one (1).
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Some other possible alternatives aren't connected to a particular setting. So, the solution is indeed the right one.

4 0
2 years ago
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