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Dmitriy789 [7]
2 years ago
11

A model rocket flies horizontally off the edge of a cliff at a velocity of 80.0m/s. If the canyon below is 128.0 m deep, how far

from the edge of the cliff does the model rocket land?
a. 112m
b. 225m
c. 337m
d. 409m
Physics
1 answer:
RSB [31]2 years ago
5 0

Answer:

c. 337

Explanation:

can someone answer my question

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C. Forces have mass and take up space
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3 years ago
what is the force per meter on a straight wire carrying 5.0 a when it is placed in a magnetic field of 0.020 t
Dvinal [7]

The Force per meter on a straight wire carrying current in a magnetic field is<u>  0.045 N/m.</u>

<u>Calculation:-</u>

       F/ℓ = B I sin θ

  Where B – Magnetic field = 0.02 T I – Current = 5 A          

Substituting the values

F/ℓ = (0.02) (5) (sin 27 deg)

F/ℓ = <u>0.045 N/m</u>

A force is an influence that can alternate the motion of an item. A force can cause an item with mass to trade its pace, i.e., to boost up. force can also be described intuitively as a push or a pull. A pressure has both value and course, making it a vector quantity.

The push or pull on an item with mass causes it to change its velocity. force is an external agent capable of converting a frame's nation of relaxation or motion. It has significance and a path. A force is a push or pulls among gadgets. it is called an interplay because if one object acts on some other, its movement is matched with the aid of a reaction from the alternative object.

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3 0
1 year ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
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Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.
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The velocity would be the same for all ramps.
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The photographer realizes that with the lens she is currently using, she can't fit the entire landscape she is trying to photogr
dexar [7]

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

What is focal length?

The focal length is a measure of how strongly the system converges or diverges light.

A positive focal length indicates that a system converges light, while a negative focal length indicates that the system diverges light.

For a standard rectilinear lens,

FOV = 2 arctan (x/2f)

FOV ∝ 1 / f

where x is the diagonal of the film.

Focal length (f) and field of view (FOV) of a lens are inversely proportional.

From the equation we can say that,

A shorter focal length gives you a wide angle of view which allows more view to fit in the frame.

Hence,

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

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