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irina [24]
3 years ago
8

A solid cylinder with a radius of 73.3 cm and a mass of 2.33 kg accelerates

Physics
1 answer:
vovangra [49]3 years ago
8 0

Answer:

(a) 20.4 rad/s

(b) 3.09 Nm

Explanation:

Given:

Δθ = 6.71 rev = 42.2 rad

ω₀ = 0 rad/s

t = 4.13 s

Find: ω

Δθ = ½ (ω + ω₀) t

42.2 rad = ½ (ω + 0 rad/s) (4.13 s)

ω = 20.4 rad/s

Find: α

Δθ = ω₀ t + ½ αt²

42.2 rad = (0 rad/s) (4.13 s) + ½ α (4.13 s)²

α = 4.94 rad/s²

∑τ = Iα

τ = ½ m r² α

τ = ½ (2.33 kg) (0.733 m)² (4.94 rad/s²)

τ = 3.09 Nm

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Two differences between science activities carried out in a science laboratory and those carried out side of science laboratory.
viva [34]

In a science lab, there is the adequate availability of raw materials for experiments such as chemicals while outside a science lab there isn't.

In a science lab, the availability of apparatuses and sources of heat is higher than outside a science lab.

<h3>What is science laboratory?</h3>

The science lab is a place where the experiments related to the science background are performed with the use of apparatus.

The science lab has plenty of equipment to perform the experiment whether related to physics or chemistry or biology.

Thus, source of heat and availability of raw material and equipment are the differences between the science activities carried out in a science laboratory and those carried out side of science laboratory.

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8 0
2 years ago
Read 2 more answers
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the
Rashid [163]

Answer:

47.8rad/s

Explanation:

For energy to be conserved.

The potential energy sustain by the object would be equal to K.E

P.E = m× g× h = 2 × 9.81× 3.5= 68.67J

Now K.E = 1/2 × I × (w1^2 - w0^2)

I = 2/3 × M × R2

= 2/3 × 2 × (0.23)^2= 0.0705

Hence

W1 = final angular velocity

Wo = initial angular velocity

From P.E = K.E we have;

68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)

(w1^2 - w0^2) = 1948.09

W1^2 = 1948.09 + (18.3^2)

W1^2=2282.98

W1 = √2282.98

=47.78rad/s

= 47.8rad/s to 1 decimal place.

7 0
3 years ago
I would appreciate some help.
motikmotik
You oughtta be able to do this one.

The "efficiency" is just the portion of the input work that
comes out in a useful form.

If the efficiency is 70%, that tells you that however much work
you put INto the machine, the machine will do 70% of that much
work for you at the output side.

Put 20,000 J in ... out comes (0.70) x (20,000 J)  =  14,000 J .

What happens to the other 30% of the work you put into it ?
It turns into HEAT.  That's why machines always have to be
cooled somehow while they're running.  
 
4 0
3 years ago
7. A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground.
miv72 [106K]

The kinetic energy, potential energy and total mechanical energy possessed by the skydriver are 1.5 × 10⁵ Joule, 6.7 × 10⁵ Joule and 8.2 × 10⁵ Joule respectively.

To find the answer, we need to know about the expression of kinetic energy and potential energy.

<h3>What are the expressions of kinetic energy and potential energy?</h3>
  • Mathematically, kinetic energy= 1/2 × mass × velocity²
  • Gravitational potential energy near the earth surface= mass × g × height on the earth surface

<h3>What's the kinetic energy, potential energy and total mechanical energy of the 78kg skydriver at 870 m on earth surface with 62 m/s velocity?</h3>
  • Kinetic energy= 1/2 × 78 × 62² = 1.5 × 10⁵ Joule
  • Potential energy= 78×9.8×870= 6.7× 10⁵ Joule
<h3>What's the total mechanical energy?</h3>
  • Mechanical energy= kinetic energy+ potential energy
  • 1.5 × 10⁵ Joule + 6.7× 10⁵ Joule = 8.2× 10⁵ Joule

Thus, we can conclude that the kinetic energy, potential energy and total mechanical energy possessed by the skydriver are 1.5 × 10⁵ Joule, 6.7 × 10⁵ Joule and 8.2 × 10⁵ Joule respectively.

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8 0
2 years ago
Antonina throws a coin straight up from a height of
vichka [17]

Answer:

s=vt-\frac{1}{2}gt^2

Explanation:

We could use the following suvat equation:

s=vt-\frac{1}{2}gt^2

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

g=-9.8 m/s^2

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s

where the negative sign means the direction is downward.

5 0
3 years ago
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