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olchik [2.2K]
2 years ago
12

An elevator held by a single cable is ascending but slowing down. Is the work done by tension positive, negative, or zero? What

about the work done by gravity? Explain
Physics
1 answer:
inessss [21]2 years ago
8 0

Explanation:

1.) Work = Force*Distance

2.) The direction of motion lies in the same direction as the tension in the cable. So the work done by tension would be positive.

3.) The direction of the weight (due to gravity) would act opposite to the direction of motion of the elevator, so work done by gravity becomes negative.

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A textbook of mass 2.09kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose di
nydimaria [60]

Answer:

(A) 9.7 N

(B) 15.4 N

 (C) I = 0.0045 kg m^{2}

Explanation:

mass of text book (M1) = 2.09 kg

mass of book (M2) = 2.99 kg

diameter of the pulley (d) = 0.12 m

radius (r) = 0.06 m

distance moved (s) = 1.30 m

time (t) = 0.75 s

acceleration due to gravity (g) = 9.8 m/s^[2}

(a) what is the tension in the part of the cord attached to the text book?

the text book is moving horizontally, so the tension in this case becomes

tension = mass x acceleration

we can get the acceleration from s = ut + 0.5 at^{2}

since the books are initially at rest u = 0

s = 0.5 at^{2}

1.3 = 0.5 x a x 0.75^{2}

a = 4.643 m/s^[2}

 

tension (T1) = 2.09 x 4.643 = 9.7 N

(b) what is the tension in the part of the cord attached to the book?

   the book is hanging vertically, so the tension in this case becomes

tension = m x ( g - a )

(g-a) is the net acceleration of the first book

tension (T2) = 2.99 x (9.8 - 4.643) = 15.4 N

(c) What is the moment of inertia of the pulley?

    if the books were to move in the direction of the book, it will cause the pulley to rotate clockwise and if they were to move in the direction of the text book on the table the pulley will rotate in an anticlockwise direction. Taking clockwise rotation of the pulley to be negative while anticlockwise to be positive, we can say  ( T2 - T1 )r = I∝

      where ∝ is the angular acceleration of the pulley relative to its radial  

      acceleration, ∝ = \frac{a}{r}

      ( T2 - T1 )r = I\frac{a}{r}

      I = \frac{(T2 - T1)r^{2}}{a}

      I = \frac{(15.5 - 9.7)0.060^{2}}{4.643}

      I = 0.0045 kg m^{2}

6 0
2 years ago
The strength of the gravitational field of a source mass can be measured by the magnitude of the acceleration due to gravity at
Svetradugi [14.3K]

Answer:

9.79211 m/s²

Explanation:

M = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6378000 m

g=G\frac{M}{r^2}\\\Rightarrow g=6.67\times 10^{-11}\frac{5.972\times 10^{24}}{(6378000)^2}\\\Rightarrow g=9.79211\ m/s^2

The acceleration due to gravity is 9.79211 m/s²

For any distance above the Earth's surface h

g=6.67\times 10^{-11}\frac{5.972\times 10^{24}}{(6378000+h)^2}\\\Rightarrow g=\frac{3.983324\times 10^{14}}{6378000+h}\ m/s^2

5 0
2 years ago
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi
morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = \dfrac{0.029158}{4.03176}= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0
1 year ago
Read 2 more answers
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit willa) decrease
slega [8]

Answer:

c) quadruple in magnitude

Explanation:

The power dissipated in the circuit is given by:

P=I^2 R

where

I is the current in the circuit

R is the total resistance of the circuit

In this problem:

- The current is doubled: I' = 2 I

- The resistance is kept constant: R' = R

So, the power dissipated is

P' = (I')^2 R' = (2I)^2 R=4 I^2 R=4 P

so, the power dissipated increase by a factor 4 (quadruples).

5 0
1 year ago
Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show
Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call R_M its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

R_M = 1000 R (1)

Both  the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

V_M = V_R

Using Ohm's law, V=RI, we can rewrite the previous equation as:

R_M I_M = R I_R

where

I_M is the current in the meter

I_R is the current in the bulb

Using (1), this equation becomes

(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}

so, the current in the meter is 1000 times less than through the bulb.

5 0
2 years ago
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