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jok3333 [9.3K]
3 years ago
5

A ball rolls off the edge of a table with a fairly large horizontal velocity. Which of the following statements are true? (Selec

t all that apply.) A.The vertical velocity is constantly increasing as the ball falls.
B.The horizontal velocity does not noticeably change as the ball falls.
C.The horizontal velocity decreases significantly as the ball falls through the air.
D.The ball will take more time to fall to the floor than if it had been dropped with no horizontal velocity.
E.The acceleration of the ball changes throughout its flight through the air. F.The ball will take less time to fall to the floor than if it had been dropped with no horizontal velocity.
G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.
H.The velocity vector of the ball changes as it travels through the air.
Physics
1 answer:
Degger [83]3 years ago
4 0

Answer:

A.The vertical velocity is constantly increasing as the ball falls.

B.The horizontal velocity does not noticeably change as the ball falls.

G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.

H.The velocity vector of the ball changes as it travels through the air.

Explanation:

As the ball is projected horizontally so here the vertical component of the velocity is zero

So the time to reach the ground is given as

H = \frac{1}{2} gt^2

so we will have

t = \sqrt{\frac{2H}{g}}

so this is the same time as the ball is dropped from H height

Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.

So overall the velocity vector will change due to net acceleration g

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Palm of your hand should be the correct answer if i remember correctly
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2 years ago
Cientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.
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Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back.

Explanation:

3 0
2 years 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
Read 2 more answers
20 points!
MariettaO [177]
Do they give answer choices? or is it free write? i’ll help if you tell me!!
8 0
2 years ago
A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat
Dafna11 [192]

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

  • To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

So

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is \sqrt{3^2+(-5)^2} =\sqrt{9+25} = \sqrt{34}

3 0
3 years ago
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