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natali 33 [55]
3 years ago
11

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

rizontal floor. Neglect any friction acting on the piano.Calculate the magnitude of the force applied by the man if he pushes
(a) parallel to the incline and
(b) parallel to the floor.

Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano (m) = 190 kg

Inclined angle (\theta) = 18 degree

Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

⇒ F=-mgsin(\theta)

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass (m) and angle (\theta) .

⇒ F=190\times 9.8\times sin(18)

⇒ F=575.38 N

b.

When the force is parallel to the floor.

⇒ Fcos(\theta)=mgsin(\theta)

⇒ F=\frac{mgsin(\theta)}{cos(\theta)}

⇒ Plugging the values.

⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

⇒ F=605 N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g
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Answer:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

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And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle \theta, given by:

f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)

Where:

Z_1 e represent the charge of the projectile (Z1=2)

Z_2 e is the target charge (z2=79)

K= 5x10^6 eV represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

t= 10^{-8] m represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

n =\frac{\rho N_A N_M}{M_g}

Where:

\rho = 19.3 g/m^3= 19300 Kg/m^3

N_A = 6.022 x10^{23} molecules/mol the Avogadro's number

N_M = 1 represent the atoms per molecule

M_g = 197 g/mol = 0.197 Kg/mol represent the molecular weigth

If we replace we got:

n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness t=10^{-8}m

And using the first formula we got:

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And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

4 0
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Answer quick!!! (Edge 2021)
mihalych1998 [28]

Answer:

I believe the answer is B.

7 0
3 years ago
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