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3 years ago

The free body diagram shows a box being pulled to the left up a 25-degree incline

Physics

2 answers:

Marysya12 [62]3 years ago

7 0

The answer according to E2020 is 45N

Greeley [361]3 years ago

5 0

The question is incomplete but still I answer to assume your thinking.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:
You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
<span>∑ F = ma
</span>As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
<span>∑ F = 0</span>
N-wcosθ=0
N= m*g*cos25°
N= m*(9.8)*(0.9063)
N= 8.8817*m
By putting the value of mass(m)(not given in the question) you will get the answer.
Hopefully, this is the answer of your question.

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

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3 years ago

Question 8 (1 point) How many protons are in an element with an atomic number of 8 and a mass number of 18? a 3 b 8 c 10 d 18

yarga [219]

Answer: B - 8

Explanation: 8 protons because number of protons is equal to number of atoms in the nucleus.

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3 years ago

. Một con lắc đơn có chiều dài dây treo 1m dao động điều hoà treo trong một xe chạy trên mặt phẳng nghiêng

lyudmila [28]

Answer:

Explanation:

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3 years ago

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IRINA_888 [86]

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3 years ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

$y=y_{0}+uSin\theta t-1/2 gt^{2}$

$y=2+20Sin5 t-4.9t^{2}$

As it hits the ground in time t, so put y = 0

$0=2+1.74 t-4.9t^{2}$

$4.9t^{2}-1.74t-2=0$

$t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}$

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

$y'=y_{0}+uSin\theta t'-1/2 gt'^{2}$

$y'=2+20Sin5 t-4.9\times0.35^{2}$

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

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2 years ago