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bearhunter [10]

3 years ago

10

Karen is sitting in a movie theater, 11 meters from the screen. The angle of elevation from her line of sight to the top of the

screen is 13°, and the angle of depression from her line of sight to the bottom of the screen is 49°. Find the height of the entire screen.

1 answer:

ZanzabumX [31]3 years ago

5 0

Answer:

15.2m

Explanation:

The height of the screen would be the sum of the elevation height and the depression height, both make a triangle with her distance to screen with their respective angles

$H = h_e + h_d$

$H = 11tan13^0 + 11tan49^0$

$H = 11(tan13^0 + tan49^0)$

$H =11*1.38 = 15.2 m$

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What is the height of a 2.00 kg object if its potential energy is 100 J?

lara31 [8.8K]

<u>Given that:</u>

height (h) = ?

mass (m) = 2 Kg ,

Potential energy (P.E.) = 100 J ;

We know that P.E = m × g × h

100 J = 2 × 9.81 × h

h = (100) ÷ (2 × 9.81)

<em> h = 5.09 m</em>

7 0

3 years ago

zepelin [54]

Answer:

t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s

Explanation:

After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s

y₀ = 2650 m

Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement

Y axis

y = y₀ + v₀ t - ½ g t²

the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{ \frac{2 y_o}{g} }$

t = √(2 2650/ 9.8)

t = 23.255 s

Therefore, for the cargo to reach the desired point, it must be launched from a distance of

x = v₀ₓ t

x = 111.76 23.255

x = 2298.98 m

at the point and arrival the speed is

vₓ = v₀ₓ = 111.76

vertical speed is

v_y = v_{oy} - gt

v_y = 0 - gt

v_y = - 9.8 23.25 555

v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement

7 0

3 years ago

1. Although mercury is a metal, it is a liquid at room temperature. Mercury melts at about -39°C. If

grigory [225]

Answer:

I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.

7 0

3 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

stira [4]

Answer:

See explanation

Explanation:

We have a mass $m$ revolving around an axis with an angular speed $\omega$ , the distance from the axis is $r$ . We are given:

$\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]$

and also the formula which states that the kinetic rotational energy of a body is:

$K =\frac{1}{2}I\omega^2$ .

Now we use the kinetic energy formula

$K =\frac{1}{2}mv^2$

where $v$ is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

$v=\omega r$

After replacing in the previous equation we get:

$K =\frac{1}{2}m(\omega r)^2$

now we have the following:

$K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2$

therefore:

$mr^2=I$

then the moment of inertia will be:

$I = 13*(0.5)^2=3.25 [Kg*m^2]$

3 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago