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Tema [17]
3 years ago
10

An elevator is being lowered at steadily decreasing speed by a steel cable attached to an electric motor. There is no air resist

ance, nor is there any friction between the elevator and the walls of the elevator shaft. Why does the upward force exerted on the elevator by the cable have a larger magnitude as the downward force of gravity on the elevator
Physics
2 answers:
shtirl [24]3 years ago
6 0

Answer:

F = W + ma  a> 0

Explanation:

For this exercise let's use Newton's second law

 we assume the upward direction as positive

          F - W = m a

          F = W + ma

          F = m (g + a)

In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

               

We can see that since a> 0 the force F must have greater than the weight of the elevator

Anvisha [2.4K]3 years ago
6 0

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • At any time, there are two forces acting on the elevator: the tension in the cable T (upward) and the force of gravity (downward).
  • Taking the upward direction as positive, if we apply Newton's 2nd law, we will have the following equation:

       T - m*g = m*a (1)

  • Now, we know that the elevator is being lowered at a decreasing speed, which means that the acceleration must have an opposite direction to the displacement.
  • Since the displacement is downward, that means that the acceleration must be positive.
  • If a ≥ 0, this means that T- mg ≥ 0, so the tension T must have a larger magnitude as the downward force of gravity on the elevator.
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7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How
Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

Y = \frac{FL}{Ae}

<u></u>

<u>Part A</u>

cross sectional area and applied force are the same as the original but the length is doubled

\frac{FL_1}{A_1e_1} =  \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} =  \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o

The second wire will be stretched by 2 mm

<u>Part B</u>

a third wire with the same length but twice the diameter of the first

\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

3 0
4 years ago
A bullet of mass 0.017 kg traveling horizontally at a high speed of 290 m/s embeds itself in a block of mass 5 kg that is sittin
rodikova [14]

Answer:

(a) vf = 0.98 m/s

(b) K₁ = 714.85 J : Total translational kinetic energy before the collision.

K₂=  2.41 J : Total translational kinetic energy after the collision.

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:    

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁ = 0.017 kg : mass of the bullet

m₂ = 5 kg : mass of the block

v₀₁ =  290 m/s : initial velocity of the bullet

v₀₂ = 0  : initial velocity of the block₂

(a) Speed of the block after the bullet embeds itself in the block

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = (m₁+ m₂)vf  

vf: final velocity of the block

( 0.017)*( 290) + (5)*(0) = ( 0.017 + 5 )*vf

4.93+ 0 = (  5.017 )*vf

vf = 4.93 / 5.017

vf = 0.98 m/s

b)  Total translational kinetic energy before (K₁) and after the collision(K₂).

K₁ = 1/2(m₁*v₀₁² + m₂*v₀₂²)

K₁ = 1/2(0.017*(290)² + 5*(0)²) = 714.85 J

K₂= 1/2(m₁+ m₂)*vf²

K₂= 1/2(0.017+ 5)*(0.98)²  = 2.41 J

6 0
3 years ago
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7 0
4 years ago
To get a sofa moving which type of friction do you need to overcome?
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Neko [114]

Answer:

Explanation:

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Work function = 2.65 eV.

Maximum kinetic energy = 4.42 - 2.65 = 1.77 eV.

8 0
4 years ago
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