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4 years ago

5

a 2-n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if a 5-n force wa

s applied?

1 answer:

Degger [83]4 years ago

3 0

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

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If a plane is inclined at an angle 5o with the horizontal and a cylindrical metal is allowed to roll between two fixed height 92

marysya [2.9K]

Answer:

0.61 second.

Explanation:

So, from the question we are given the following useful parameters or data or information that is going to assist us in solving this question and they are;

(1). Inclination of plane, θ = 50°.

(2). ''The horizontal and a cylindrical metal is allowed to roll between two fixed height 92cm apart on the plane" that is height = 0.92 m(92cm).

(3). The "acceleration of the cylinder to be 2/3 × g Sin (theta) and g = Acceleration due to gravity = 9.8m/s^2

Soz let us delve right into the solution to the question above. We will be making use of the mathematical representation or Equation below;

s = (1/2) a × t².

Where s = height, a = acceleration due to gravity and t = time.

0.92 = (9.8 × sin 50°/ 3) × t².

t² = 0.37

t = √ 0.37.

t =0.61 seconds.

8 0

4 years ago

Arte-miy333 [17]

Explanation:

Pressure = force / area

P = (68 kg × 9.8 m/s²) / (2 × (0.04 m)²)

P = 208,250 Pa

Converting to psi:

P = 208,250 N/m² × (0.225 lbf/N) × (0.0254 m/in)²

P = 30.2 psi

4 0

4 years ago

The physical structure of the earth’s rock is changed by _____.

PSYCHO15rus [73]

It could be stress or strain

4 0

4 years ago

Read 2 more answers

The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/

svp [43]

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is $\frac{1}{6}$

So, the equation is

$\frac{1}{6} = 1 - \frac{T_2}{T_1}$

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is $\frac{1}{3}$

Therefore the next equation is

$\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}$

Now solve both the equations

solve equations (1) and (2)

$2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)$

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

7 0

4 years ago

The drawing shows a tire of radius R on a moving car

masha68 [24]

Answer:

H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

Em₀ = K = 1/2 m v²

Final. In the sought height

$Em_{f}$ = U = mg h

Energy is conserved

Em₀ = $Em_{f}$

½ m v² = m g h

v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

Em₀ = K = ½ I w²

Final. Heights sought

Emf = U = m g R

Em₀ = $Em_{f}$

½ I w² = m g R

The moment of inertial of a cylinder is

I = $I_{cm}$ + ½ m R²

I= ½ $I_{cm}$ + ½ m R²

Linear and rotational speed are related

v = w / R

w = v / R

We replace

½ $I_{cm}$ w² + ½ m R² w² = m g R

moment of inertia of the center of mass

$I_{cm}$ = ½ m R²

½ ½ m R² (v²/R²) + ½ m v² = m gR

m v² ( ¼ + ½ ) = m g R

v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

2 g H = 4/3 g R

H / R = 2/3

7 0

3 years ago