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2 years ago

One example of an energy-efficient device is a(n)

Physics

2 answers:

Hunter-Best [27]2 years ago

7 0

A. Fluorescent light bulb

Rom4ik [11]2 years ago

5 0

A fluorescent light bulb is the most energy efficient in this group.

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4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C

aniked [119]

Answer:

a) ΔL/L = F / (E A), b) $L_{f}$ = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

P = E ΔL / L

The pressure is defined as

P = F / A

Let's replace

F / A = E ΔL / L

F = E A ΔL / L

ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

ΔL = L [F / EA]

$L_{f}$ -L = L (F / EA)

$L_{f}$ = L + L (F / EA)

$L_{f}$ = L (1 + L (F / EA))

4 0

3 years ago

What is the value of y?<br>18<br>10<br>2y + 4<br>10 + 2х

DENIUS [597]

Answer:

can u please write it correct like sorry I dont understand it

4 0

3 years ago

Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7

Dmitry [639]

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

$v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}$ ......(I)

The orbital speed of satellite B,

$v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}$ ......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}$

Put the value into the formula

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}$

$\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}$

Hence, The ratio of their orbital speeds are 5:4.

8 0

2 years ago

Where do tadpoles live? a. on land c. both on land and water b. in water d. none of the above

Bezzdna [24]

Tadpoles live in water

5 0

2 years ago

Read 2 more answers

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago