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Alinara [238K]
2 years ago
4

I solid iron ball of mass 770kg is used on a building site. The ball is suspended by a rope from a crane. The distance from the

point of suspension to the centre of the mass of the ball is 12 m.The ball is pulled back from the vertical and then realeased, it falls through a vertical height of 1.6m and strikes the wall: What is the speed of the ball just before impact? ...?
Physics
1 answer:
Hoochie [10]2 years ago
3 0
Based on the given values above, here is the solution which I hope will be helpful to you. 
Triangles, L = 12m, x_0 = 1.6,
we need to find the angle (theta)
sin (theta) = 1.6/12 = 0.1333....
theta = ArcSin(0.1333...) = 0.1337 rad
 hen, this is the height that the mass vertically raises in it's arc
y_2 = L-L*cos(theta) = 0.107 m
use y_2 in a kinematic swing... v=√2gy^2 = 1.45m/s
The speed would be 1.45m/s. Hope this answer helps.
You might be interested in
carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr
denis23 [38]

The half-life equation m=m_{0} (\frac{1}{2})^n in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>

<em>n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}</em>

<em>\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2</em>

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

<em>2 * 5700 =  </em>11400 yr

3 0
2 years ago
Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener
Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 \frac{m^{3} }{kg}

% Error =  20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 \frac{m^{3} }{kg}

% Error =  - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 \frac{J}{kg k}

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ \frac{V}{m} = \frac{R T}{P}

⇒ Here \frac{V}{m} = Specific volume = v

⇒ v =  \frac{R T}{P}

Put all the values in the above formula we get

⇒ v = \frac{323}{10^{6} } ×81.49

⇒ v = 0.02632 \frac{m^{3} }{kg}

This is the specific volume by ideal gas equation.

Actual value = 0.021796 \frac{m^{3} }{kg}

Error =  0.02632 - 0.021796 =   0.004524 \frac{m^{3} }{kg}

% Error =  \frac{0.004524}{0.021796} × 100

% Error =  20.75 %

(b). From the generalized compressibility chart the value of specific volume

 \frac{V}{m} = v = 0.0142 \frac{m^{3} }{kg}

The actual value = 0.021796 \frac{m^{3} }{kg}

Error = 0.0142 - 0.021796 =  \frac{m^{3} }{kg}

% Error = \frac{- 0.0076}{0.021796} × 100

% Error =  - 34.85 %

3 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
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
2 years ago
When Jane drives to work, she always places her purse on the passenger’s seat. By the time she gets to work, her purse has falle
LUCKY_DIMON [66]
At some time during her drive she backed up with a substantial negative. ( backwards) acceleration. Since the pocket book is not physically connected to the seat it is free to move. Upon rapid negative acceleration the pocket book remains in its position while the car accelerates backwards away from it. this demonstrates Newtons 1st law of motion. The first law is the law of inertia. Which states, an object at rest. ( pocketbook) will remain at rest and an object in motion will continue in motion at constant velocity, unless acted upon by some outside force to change its motion.
4 0
2 years ago
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