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Alika [10]
3 years ago
9

Calculate the gravitational potential energy of 3kg book sitting on shelf 4m high

Physics
2 answers:
n200080 [17]3 years ago
7 0

Answer:

nope loser

Explanation:

likoan [24]3 years ago
5 0

Answer:

i know it i know it pick me

Explanation:

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If atmospheric pressure on a certain day is 749 mmhg, what is the partial pressure of nitrogen, given that nitrogen is about 78%
VARVARA [1.3K]
The partial pressure of nitrogen is about 584 mmHg.
7 0
3 years ago
A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0
2 years ago
in addition to studies of the body system of corals what studies in earth science might increased scientists understanding of an
Maru [420]

Answer: they could get a piece of the coral and run tests on it

Explanation:

6 0
3 years ago
motorcycle accelerates from rest to a final velocity of 27 m/s in 9 seconds. What is its acceleration?
Vedmedyk [2.9K]

Answer:

3 m/s squared

Explanation:

The formula you use is Vf= Vi + at. You rearrange it to a= Vf - Vi/t. The Vf is 27m/s. The Vi is 0m/s and the t is 9s. Cross out Vi since it’s zero and you’re left with a= 27m/s divided by 9s, which equals 3

5 0
3 years ago
A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o
Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

KE_i=\dfrac{1}{2}mu^2

KE_i=\dfrac{1}{2}0.3\times 10^2

KEi= 15 J

Final kinetic energy

KE_f=\dfrac{1}{2}mv^2

KE_f=\dfrac{1}{2}0.3\times 5^2

KEf=3.75 J

The  energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

 V=\sqrt{r.g}

Now by putting the values

V=\sqrt{0.8\times 10}

V= 2.82 m/s

So kinetic energy KE

KE=\dfrac{1}{2}mV^2

KE=\dfrac{1}{2}0.3\times 2.82^2

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

                                                                                                    = 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0
3 years ago
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