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astraxan [27]
2 years ago
14

A hobby rocket reaches a height of 72.3 meters and lands 111 meters from the launch point

Physics
1 answer:
faust18 [17]2 years ago
8 0
Using the following formulas for projectile motion:

Height, H = ( Vo^2 * sin theta^2 )/g

Range, R = ( Vo^2 * sin 2*theta )/g

Rearranging in terms of Vo^2:

Vo^2 = gH / sin theta^2

Vo^2 = gR / sin 2*theta

Equating the two formulas to each other to solve for the angle theta:

gR / sin 2*theta = <span>gH / sin theta^2
</span>
Substituting the given values:

(9.8)(111) / sin 2*theta = (9.8)(72.3)<span> / sin theta^2
</span>angle = 52.36 degrees

Therefore, the angle of launch is approximately 52.36 degrees.
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At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta
Lubov Fominskaja [6]

Answer:

a=2330

b= 0.223secs

Explanation:

pb=2330m

t=0.223secs

6 0
2 years ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
2 years ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

      v = v₀ - a t

     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

    a ’= (v₀-v) / 2T

Let's substitute the land values ​​in plot X

     a’= a / 2

Now let's use Newton's second law

       W = ma

      m g = m a

      a = g

We substitute

      a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0
1 year ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
Alja [10]

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s    

7 0
1 year ago
Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the
Andre45 [30]

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, E=3\times 10^6\ V/m

Let the electrical potential is, V=10^8\ V

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

5 0
2 years ago
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