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Phoenix [80]
3 years ago
10

One of your fellow students comes up to you and asks the following question, "If an object is not moving, can it be accelerating

?" Based on your prior knowledge of this topic, what is the best response? A). Yes, becuase acceleration depends on the rate of change of velocity and not the value of velocity itself.
B). No, acceleration and velocity are equivalent, so if one equals zero, the other must also equal zero.
D). Yes, but only if the object's acceleration is very small.
D). Yes, but only if the acceleration is negative in order to keep velocity equal to zero.
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

correct answer is option A (Yes, becuase acceleration depends on the rate of change of velocity and not the value of velocity itself.)

Explanation:

we know that from first equation of motion,

v=u+at

0r

a=\frac{v-u}{t}............(1)

here,

v-final velocity

u-initial velocity

a-acceleration

t-time

if a body is not moving it means if its initial velocity is zero (u=0) and if after time t it moves with velocity v then its acceleration will be given as

a=\frac{v}{t}...........(2)

From above equation it is clear that if an object is not moving, it can be accelerating because acceleration depends on rate of change of velocity and not the value of velocity itself.

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A soccer ball is released from rest at the top of a grassy incline. After 8.6 seconds, the ball travels 87 meters and 1.0 s afte
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A batter hits a softball over a third baseman's head with speed v0 and at an angle ?from the horizontal. Immediately after the b
anastassius [24]

Answer:

a) The magnitude of the initial velocity is 18.77 m/s.

b) The launching angle is 31.51°.

c) The horizontal component of the velocity at t = 1.900 s is 16.00 m/s.

The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.

d) The horizontal component of the position vector at time t = 1.900 s is 30.40 m.

The vertical component of the position vector at time t = 1.900 s is 0.9375 m

Explanation:

Hi there!

The equations for the velocity and position vector of the ball are the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity.

v = velocity vector at time t.

a and b) First, let´s find the range of the ball, i.e. the horizontal distance traveled by the ball.

The distance traveled by the baseman can be calculated with this equation:

x = v · t

Where:

x =traveled distance.

v = velocity.

t = time

Then:

x = 7.000 m/s · 2.000 s

x = 14.00 m

The baseman runs 14.00 m. Since he was located 18.00 from the home plate, the horizontal distance traveled by the ball is (14.00 m + 18.00 m) 32.00 m.

If we locate the origin of the frame of reference at the point where the ball is hit, the initial vertical and horizontal positions (x0 and y0) are zero. Since the ball is caught at the same height at which it left the bat, the vertical position of the ball when it is caught is 0.

So, the position vector of the ball at the time when it is caught (2 s after it is hit), is the following:

r = (32.00 m, 0 m)

Using the equations of the x- and y-components of the position vector, we can obtain the initial velocity and the angle:

rx = x0 + v0 · t · cos α     (x0 = 0)

ry = y0 + v0 · t · sin α + 1/2 · g · t²         (y0 = 0)

rx = 32.00 m = v0 · 2.000 s · cos α

ry = 0 m = v0 · 2.000 s · sin α - 1/2 · 9.807 m/s² · (2.000 s)²

Solving the first equation for v0:

16.00 m/s / cos α = v0

And replacing v0 in the second equation:

0 m = 32 m · sin α / cos α - 1/2 · 9.807 m/s² · (2.000 s)²

1/2 · 9.807 m/s² · (2.000 s)² = 32 m · tan α

1/2 · 9.807 m/s² · (2.000 s)² / 32 m = tan α

α = 31.51°

<u>b) The launching angle is 31.51°</u>

The initial velocity will be:

16.00 m/s / cos α = v0

16.00 m/s / cos (31.51°) = v0

v0 = 18.77 m/s

<u>a) The magnitude of the initial velocity is 18.77 m/s.</u>

<u />

c) Let´s use the equation of the velocity vector:

v = (v0 · cos α, v0 · sin α + g · t)

vx = v0 · cos α

vy = v0 · sin α + g · t

The horizontal component of the velocity does not depend on time (neglecting air resistance).

Then:

vx = 18.77 m/s · cos (31.51°)

vx = 16.00 m/s

<u />

<u>0.100 s before the ball is caught, the horizontal component of the velocity is 16.00 m/s. </u>

Now let´s calculate the vertical component of the velocity:

vy = 18.77 m/s · sin (31.51°) - 9.807 m/s² · 1.900 s

vy = -8.823 m/s

<u>The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.</u>

d) Let´s use the same equations we have used in part a).

x = x0 + v0 · t · cos α

x = 18.77 m/s · 1.900 s · cos (31.51°)

x = 30.40 m

<u>The horizontal component of the position vector at time t = 1.900 s is 30.40 m</u>

<u />

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 18.77 m/s · 1.900 s · sin (31.51°) - 1/2 · 9.807 m/s² · (1.900 s)²

y = 0.9375 m

<u>The vertical component of the position vector at time t = 1.900 s is 0.9375 m </u>

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The collector current is directly proportional to the base current
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Answer:

As the rock falls from the edge of the cliff, gravitational potential energy is converted to mechanical kinetic energy.

Explanation:

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