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2 years ago

8. What is the mass of an object if a force of 34 N produces an acceleration of 4 m/s/s?

Physics

2 answers:

Oxana [17]2 years ago

8 0

Answer:

hello i need brain points sorry for not answering the question

Explanation:

coldgirl [10]2 years ago

7 0

Answer:

<h3>The answer is 8.5 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

where

f is the force

a is the acceleration

So we have

$m = \frac{34}{4} = \frac{17}{2} \\$

We have the final answer as

Hope this helps you

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A car starts from rest and accelerates uniformly over a time of 7.25 seconds for a distance of 210 m. Determine the acceleration

lara31 [8.8K]

Answer:

Acceleration is 7.990487515m/s²

Initial velocity is 0m.s

Explanation:

s=ut+(1/2)at²

210=0(7.25)+(1/2)a(7.25²)

210=26.28125a

∴a=7.990487515m/s²

'Vi' or 'u' is the inital speed. Since it starts from rest, this equals 0.

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1 year ago

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Inessa05 [86]

The moon lacks an atmosphere compared to the Earth. Hope this helps!

4 0

2 years ago

Read 2 more answers

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

5 0

2 years ago

Answer the question in French:

frez [133]

Bonjour

C'est la fille de ma mère mais ce n'est pas moi, qui est-ce ?

<em>She is my mother daughter but, she's not me ? Who is she ?</em>

<em />

<em>Ma sœur </em><em> = my sister</em>

<em />

6 0

2 years ago

the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar

adell [148]

Answer:

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

so it is given by

specific gravity = weight of the object/weight of the water displaced

now we have

weight of the object = (density)(volume)g

weight of object = (19.3)(0.55)g

now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3)(0.55ml)g

now we have

specific gravity = (19.3×0.55)/(1×0.55)

specific gravity= 19.3

8 0

2 years ago