Are active centers of galaxies that radiate powerful x-rays visible across the universe

A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig

Blababa [14]

Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a) If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf ⇒ I₁ * ω₁ = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)

W = 7.03 J

4 0

3 years ago

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

$\sf \implies s = 100 \: m$

6 0

4 years ago

the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes

slamgirl [31]

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.

5 0

3 years ago

Compare two processes we use to measure the age of Earth and its parts.

yan [13]

Answer:

Generally speaking, scientists have developed four different methods of determining the age of the earth. By using these methods, or a combination of them, the age of geological formations created by past events and even the fossilized bones of prehistoric animals can be determined.

Based on the research done by scientists, the Geologic Column, a graph that illustrates earth age, was developed. The Column, with its names for the epochs and eras of time, illustrates what scientists think was taking place in earth history. A small version of the Geologic Column is pictured at the right.

Explanation:

I hope this helps :)

3 0

3 years ago

A 5.5-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the

Stella [2.4K]

Answer:

16.5 kwh and 59400 kJ.

Explanation:

kWh is a measure of energy that is equivalent to the power in kw times the number of hours the device worked.

In this case, it would be equal to:

$E_{kwh}=5.5kw*3h=16.5kwh$

1 kw also means 1kj of energy spent per second. With this, we calculate the amount of energy in kJ spent by the resistance:

$E_{kJ}=5.5\frac{kJ}{s}*3h*\frac{3600s}{1h} = 59400 kJ$

5 0

3 years ago