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3 years ago

Two students make the following claims:

Physics

1 answer:

antiseptic1488 [7]3 years ago

6 0

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

$F=m.g$

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

$W=F.s\cos\theta$

where:

$s=$ displacement of the object

$\theta=$ angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

In case of free-fall the angle between the force is and the displacement is zero.
In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

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You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

Radda [10]

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

6 0

2 years ago

While playing a scavenger hunt game, Anthony walks 48.0 m to the south and then walks 12.0 m to the west. What single straight-l

Alenkasestr [34]

As these are distances created by moving in a straight line, using a trigonometric analysis can solve the missing single straight-line displacement. Looking at the 48m and 12m movements as legs of a triangle, obtaining the hypotenuse using the pythagorean theorem will yield us the correct answer.
This is shown below:

c^2 = 48^2 + 12^2
c = sqrt(2304 + 144)
c = sqrt(2448)
c = 49.48 m

To obtain the angle at which Anthony walks 49.48, we obtain the arc tangent of (12/48). This is shown below:

arc tan (12/48) =14.04 degrees.

Therefore, Anthony could have walked 49.48 m towards the S 14.04 W direction.

4 0

2 years ago

HEY CAN ANYONE PLS HELP ME OUT IN DIS PLS

Zanzabum

Answer:

Its the second choice

Explanation:

4 0

2 years ago

Read 2 more answers

A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp

Sergeu [11.5K]

Answer:

$U=1.29\times 10^{-7}\ J$

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm ,K₁= 4.7

d₂=2.1 mm , K₂=2.6

We know that capacitance given as

$C_1=\dfrac{K_1\varepsilon _oA}{d_1}$

$C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_1=1.26\times 10^{-10}\ F$

$C_2=\dfrac{K_2\varepsilon _oA}{d_2}$

$C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_2=0.701\times 10^{-10}\ F$

Net capacitance

$C=\dfrac{C_1C_2}{C_1+C_2}$

$C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F$

$C=4.5\times 10^{-11}\ F$

We know that stored energy given as

$U=\dfrac{CV^2}{2}$

V= 76 V

$U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J$

$U=1.29\times 10^{-7}\ J$

3 0

2 years ago

A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a n

Naddik [55]

Answer:

1917723.40119 m/s

Explanation:

m = Mass of deuteron = $(1.637+1.675)\times 10^{-27}=3.312\times 10^{-27}$

k = Boltzmann constant = $1.381\times 10^{-23}\ J/K$

T = Temperature = $2.94\times 10^8\ K$

RMS velocity is given by

$V_r=\sqrt{\dfrac{3kT}{m}}\\\Rightarrow V_r=\sqrt{\dfrac{3\times 1.381\times 10^{-23}\times 2.94\times 10^8}{3.312\times 10^{-27}}}\\\Rightarrow V_r=1917723.40119\ m/s$

The RMS velocity of the deutrons is 1917723.40119 m/s

8 0

3 years ago