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3 years ago

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Two forces act on a 8.50-kg object. One of the forces is 14.0 N. If the object accelerates at 3.50 m/s2 , what is the greatest p

ossible magnitude of the other force

1 answer:

astra-53 [7]3 years ago

7 0

Answer:

The magnitude of the other force is 43.75 N.

Explanation:

Given that,

Mass of the object, m = 8.5 kg

Force 1, $F_1=14\ N$

Acceleration of the object, $a=3.5\ m/s^2$

To find,

The greatest possible magnitude of the other force.

Solution,

Let $F_2$ is the magnitude of other force that is acting on the object. For the greatest force, the two forces must be act in opposite direction such that :

$F_2-F_1=ma$

$F_2=ma+F_1$

$F_2=14+8.5\times 3.5$

$F_2=43.75\ N$

So, the magnitude of the other force is 43.75 N.

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Answer:

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Solution:

As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

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Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = $\pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}$

Resistivity is given by:

$\rho = R\frac{A}{l}$

$R = \rho \frac{l}{A}$

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$V = 0.1\times 1210.69 = 121.069\ V$

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Grams is another unit for mass.

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An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . Suppose that

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Answer:

0.0241875 m

Explanation:

$m_1$ = Mass of quarterback = 80 kg

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Time taken = 0.3 seconds

In this system as the linear momentum is conserved

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s$

Assuming this velocity is constant

$Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m$

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2 years ago

Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas

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Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = $9.232 g/cm^3$

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}$

$a = 3.35\times 10^{-8} cm$

Atomic radius of the polonium in unit cell = r

r = 0.5a

$r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm$

$1 cm = 10^{10} pm$

$1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm$

Approximate atomic radius for polonium-209 is 167.5 pm.

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3 years ago