What is the purpose of the overrunning clutch in the starter drive?

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Reil [10]

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

6 0

3 years ago

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WHAT WILL BE YOUR CONCLUSI0N ABOUT ELECTROMAGNETIC WAVE

andrezito [222]

Answer:

It's really important waves type for fulfill human's needs

Explanation:

Electromagnetic waves are transverse waves composed by the perpendicular oscillating electric and magnetic fields.

EM waves have both Electrical and magnetic features.

they travel in the velocity of light (3*10⁸ ms⁻¹)

they does not require any media to travel. It has two perpendicular electric field and the magnetic field which are perpendicular to each other

They travel perpendicular to each of those electric and magnetic fields.

Example :

Radio Wave
Micro Wave
IR wave
Light Wave
UV rays
X rays
Gamma rays
Cosmic rays

The main importance of em waves is they allow energy to be stored within them and then can be propagated over a large distance using the dielectric and magnetic properties of materials .

This performance has been used in many fields wisely and effectively to make the things easy.

Ex : medicine , Telecommunication , energy , Engineering etc

8 0

3 years ago

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m

Gelneren [198K]

Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

a.

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

= 4.98 × 10¹⁹ protons

≅ 5 × 10¹⁹ protons

b

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

= 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

= 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

= 30 × 10⁷/14.62

= 2.05 × 10⁷ °C

5 0

3 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

6 0

3 years ago

In order to design an experiment you need a ____ about the scientific question you are trying to answer

s2008m [1.1K]

C Hypothesis. It is the correct answer.

6 0

3 years ago

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