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ValentinkaMS [17]
4 years ago
11

Determine the amount of potential energy of a 5 newton book that is moved to three different shelves on a bookcase. The height o

f each shelf is 1.0 meter, 1.5 meters, and 2.0 meters.
Physics
1 answer:
BigorU [14]4 years ago
5 0
Gravitational Potential Energy = weight x height

for 1 meter:

GPE = 5 x 1
 = <u>5N</u>

for 1.5 metres:

GPE = 5 x 1.5
 = <u>7.5N</u>

for 2 metres:

GPE = 5 x 2
 = <u>10N</u>
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In order to be considered work, the components that must be present are
atroni [7]
The components that must be present for work to be considered is a force and a movement in the same direction as the force. In the basic definition of work, a magnitude and displacement that occurs in the same direction is what makes up work. Among the choices, the correct answer is the first one.
6 0
3 years ago
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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav
uranmaximum [27]

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}

Where n = 0, 1, 2, 3, ...

Making n+1 = n

\lambda=\dfrac{2L}{n}

where n = 1, 2, 3 .....

For fundamental frequency n = 1

\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m

Frequency is given by

f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz

The fundamental frequency is 68.8 Hz

First overtone

2f=2\times 68.8=137.6\ Hz

Second overtone

3f=3\times 68.8=206.4\ Hz

The overtones are 137.6 Hz, 206.4 Hz

4 0
4 years ago
The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th
Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec,  B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>

at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 =  180 rad/sec

<u>b) Determine the force exerted on B2 at P</u>

T2 = mB1g + T1  -------- ( 1 )

where ; T1 = mB2g  ( at point p )

                 = 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

<u />

3 0
3 years ago
6. Find the magnetic field strength at the centre of a solenoid with
inysia [295]

Answer:

0.03 T

Explanation:

The magnetic field B at the center of a solenoid is given by B = μ₀Ni/L where

μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, N = number of turns of solenoid = 5000 turns, i = current in solenoid = 5 A and L= length of solenoid. Since we are not given length of solenoid, let us assume it is 1 meter. So, L = 1 m

So, B = μ₀Ni/L

=  4π × 10⁻⁷ H/m × 5000 turns × 5 A/1m

= 4π × 10⁻⁷ H/m × 25000 A-turns/m

= 314159.27 × 10⁻⁷ T

= 0.031415927 T

≅ 0.03 T

8 0
3 years ago
When discharging a fire extinguisher, it is important to aim: a. On top of the flames b. At the top of the flames c. Anywhere d.
Mumz [18]

Answer:

d. At the base of the fire

Explanation:

Most fire extinguisher utilize the non-combustible nature of carbon-dioxide . carbon-dioxide is what is used in fire extinguishers, because it does not support burning. Also, carbon-dioxide is denser than air, allowing it to sink down. Aiming at the bottom of the flame allows the carbon-dioxide to sink down on the base of the flame, covering the base of the fire, and cutting off its supply of oxygen which is vital for combustion.

4 0
3 years ago
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