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White raven [17]
3 years ago
15

If it takes 150 n of force to accelerate an object at 30 m/s what is the mass of the object

Physics
1 answer:
iris [78.8K]3 years ago
3 0
F  =  ma.         

Note Force should  be = 150 N.  Acceleration = 30 m/s2   ( Am presuming for your question you meant to write 30m/s2  and not 30m/s as you wrote)

150 = m ( 30)

150/30  = m.

5 = m

Mass of the object = 5 Kg.
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Find the intensity of a 55 dB sound given I 0=10^-12W/m^2
MakcuM [25]

Answer:

3.16 × 10^{-7} W/m^{2}

Explanation:

β(dB)=10 × log_{10}(\frac{I}{I_{0} })

I_{0}=10^{-12} W/m^{2}

β=55 dB

Therefore plugging into the equation the values,

55=10 log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

5.5= log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

10^{5.5}= \frac{I}{10^{-12} }

316227.76×10^{-12}= I

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2 years ago
What is a suspension bridge- in your own words please and ty
PIT_PIT [208]
A bridge supported by vertical cables which then leads to more support from larger cables.
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3 years ago
Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr
Stolb23 [73]

Answer:

2.464 cm above the water surface

Explanation:

Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

weight of the block = 0.78 * 11.2^3  gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

6 0
3 years ago
A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

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Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

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Hence, P_{max} will be calculated as follows.

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Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
3 years ago
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To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

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Here,

h = Depth average

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Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

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F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

6 0
2 years ago
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