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4 years ago

8

Each of the following objects possesses energy. Which forms of energy are mechanical, which are nonmechanical, and which are a c

ombination?
(a) glowing embers in a campfire
(b) a strong wind
(c) a swinging pendulum
(d) a person sitting on a mattress
(e) a rocket being launched into space

1 answer:

nydimaria [60]4 years ago

7 0

Explanation:

a) glowing embers in a campfire : It is a non mechanical conversion

(b) a strong wind: Kinetic or mechanical

(c) a swinging pendulum: A combination of potential and kinetic energy.

(d) a person sitting on a mattress: potential energy or mechanical

(e) a rocket being launched into space : kinetic and potential.

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A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

lara [203]

Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

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4 years ago

A type of energy work that utilizes touch and visualization

nydimaria [60]

It would be the <span>Energy Therapies Reiki that is a common type of work wherein it makes use of touch and visualisation. In addition, this kind of therapy has its primary goal just to lessen stress and furthermore promotes relaxation. The technique is said to be making use of a "life force energy."</span>

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3 years ago

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

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3 years ago

a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

Mademuasel [1]

We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
$T_1 - W_p - W_s + T_2 =0$
where
$W_p = (90kg)(9.81 m/s^2)=883 N$ is the weight of the person
$W_s = (200kg)(9.81 m/s^2)=1962 N$ is the weight of the scaffold
Re-arranging, we can write the equation as
$T_1 = 2845 N-T_2$ (1)

2) Equilibrium of torques:
$T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0$
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using $W_p = 883 N$ and replacing T1 with (1), we find
$2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0$
from which we find
$T_2 = 1128 N$

And then, substituting T2 into (1), we find
$T_1 = 1717 N$

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3 years ago

Determine (a) the starting height, (b) the time to hit the ground, and (c) the velocity when it hits the ground for an object sh

Salsk061 [2.6K]

Answer:

a

Explanation:

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3 years ago