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n200080 [17]
3 years ago
12

Layla, whose mass is 55 kg, is in an elevator that is accelerating downward at 0.6 m/s2. Layla's apparent weight is? ​

Physics
1 answer:
leva [86]3 years ago
7 0

Answer:

Explanation:

25318349

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A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
A box with a mass of 12.5kg sits on the floor how high would you need to lift it has a GPE of 568j
Degger [83]
GPE=mgh
m= 12.5kg
g= 9.81 always
h=?

568=12.5*9.81*h
Solve for h
You will get 4.63m
4 0
3 years ago
Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravita
12345 [234]

Answer:

6.49 x 10^-8 N

Explanation:

formula is

F= G * ((m1 * m2)/r^2)

F = 6.67x10^-11 * ((6.8*6.8/.218)

F = 6.49 x 10^-8 Newtons

3 0
3 years ago
Materials that are soft and porous will absorb energy causing a decrease in amplitude and energy of the sound. This is called __
Vedmedyk [2.9K]
Damping of the sound
5 0
3 years ago
An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2010 × 103 seconds (about
vazorg [7]

Answer:

Radial acceleration of moon is a_{r} = 2.246\times 10^{-3}\frac{m}{s^{2} }

Explanation:

Given :

Time period T = 1.987 \times 10^{6} sec

Distance from center of moon to planet r = 225 \times 10^{6} m

From the equation of radial acceleration,

  a_{r} = r\omega ^{2}

Where \omega = 2\pi f = \frac{2\pi }{T}

So   \omega = 3.16 \times 10^{-6} \frac{rad}{s}

Now moon's radial acceleration,

 a_{r} = 225 \times 10^{6} \times (3.16 \times 10^{-6} )^{2}

 a_{r} = 2246.76 \times 10^{-6}

 a_{r} = 2.246\times 10^{-3} \frac{m}{s^{2} }

7 0
3 years ago
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