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Ilya [14]
2 years ago
14

The two hot-air balloons in the drawing are 48.2 and 61.0 m above the ground. A person in the left balloon observes that the rig

ht balloon is 13.3° above the horizontal. What is the horizontal distance x between the two balloons?

Physics
1 answer:
Lynna [10]2 years ago
8 0

Answer:

x = 54.15 meters between the balloons horizontally.

Explanation:

Hence, vertical distance between balloons is 12.8 meters.

y= 61-48.2  = 12.2 m

A right-angle triangle is formed with,

angle θ = 13.3 degrees

opposite (y) = 12.8

adjacent (x) = ?

tan\theta = \dfrac{y}{x}

Now by putting the values in the above equation.

tan13^{\circ}=\dfrac{12.8}{x}

x=\dfrac{12.8}{tan13^{\circ}}

x = 54.15 meters between the balloons horizontally.

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xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

7 0
2 years ago
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag
Afina-wow [57]

Answer:

The induced emf in the loop is 7.35\times 10^{-4}\ V

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, A=a^2=(0.305)^2=0.0930\ m^2

Circumference of the loop,

C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m

Area of circle,

A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2

The induced emf is given by :

\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V

So, the induced emf in the loop is 7.35\times 10^{-4}\ V

8 0
2 years ago
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7 0
3 years ago
Read 2 more answers
The tops of the towers of the Golden Gate Bridge in San franciscoo , are 227 m above the water. Suppose a worker drops a 655 g w
yaroslaw [1]

Answer:

Kinetic energy is 1425.11 J.

Explanation:

Given:

Mass of the wrench is, m=655\ g=0.655\ kg

Height of fall is, h=227\ m

Force of resistance is, F=0.141\ N

Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.

Now, potential energy at the top is given as:

U=mgh\\U=0.655\times 9.8\times 227\\U=1457.113\ J

Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.

Potential Energy = Kinetic energy + Energy to overcome resistance.

⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.

Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:

W_{res}=Fh=0.141\times 227=32.007\ J

Therefore, Kinetic energy at the bottom is given as:

K=U-W_{res}\\\\K=1457.113\ J - 32.007\ J\\\\K=1425.106\approx1425.11\ J

Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.

7 0
3 years ago
A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3
vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

k = 5\:\text{N/m}

The spring potential energy PE_s is

PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2

\:\:\:\:\:\:\:=22.5\:\text{J}

3 0
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