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Pie
3 years ago
12

A solid iron ball of mass 770 kg 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 mass of the ball is 12 m.
Physics
1 answer:
Scrat [10]3 years ago
6 0
From a balistics pendulum as an example, which is probably where you are at... 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 Then, 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... <span><span>v=sqrt(<span><span>2g<span>y_2)</span></span></span>=1.45m/s</span></span>
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Carbon-14 is used to determine the time an organism was living. The amount of carbon-14 an organism has is constant with the atm
lutik1710 [3]

Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} } (1)

Where:

n_{o} - Initial amount of carbon-14.

n(t) - Current amount of carbon-14.

t - Time, measured in years.

\tau - Time constant, measured in years.

Then, we clear the time within the formula:

t = -\tau \cdot \ln \frac{n(t)}{n_{o}} (2)

In addition, time constant can be calculated by means of half-life of carbon-14 (t_{1/2}), measured in years:

\tau = \frac{t_{1/2}}{\ln 2}

If we know that \frac{n(t)}{n_{o}} = 0.25 and t_{1/2} = 5730\,yr, then the age of the organism is:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.25

t \approx 11460.001\,yr

The age of the organism is approximately 11460 years.

8 0
3 years ago
Read 2 more answers
Which is the correct formula for cellular respiration?
Tanya [424]

D (Glucose +Oxygen --> Carbon Dioxide + Water + Energy)

8 0
3 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0
3 years ago
Write a question you would like to see on a quiz to calculate NET FORCE.
Pie

Answer:

How do you find the net force?

Explanation:

i think this is want you asked for?

5 0
2 years ago
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0
3 years ago
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