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3 years ago

A ball moving with an initial velocity of 5 m/s comes to rest after 2s. What was the ball's acceleration?

Physics

1 answer:

Inga [223]3 years ago

6 0

Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

a = -2.5m/s²

Therefore, the acceleration of the ball is -2.5m/s²

NB: The negative sign shows that the ball was actually decelerating.

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2 years ago

A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the

Julli [10]

Answer:

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0

3 years ago

Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t

LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

$x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}$

Substituting into the equation, we have;

$x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}$

$x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}$

$x = \frac {-32 \pm \sqrt {1024 + 800}}{2}$

$x = \frac {-32 \pm \sqrt {1824}}{2}$

$x = \frac {-32 \pm 42.71}{2}$

$x_{1} = \frac {-32 + 42.71}{2}$

$x_{1} = \frac {10.71}{2}$

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

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3 years ago

How fast, in rpm, would a 5.8 kg, 22-cm-diameter bowling ball have to spin to have an angular momentum of 0.21 kgm2/s

-Dominant- [34]

Answer:

91 rpm

Explanation:

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3 years ago

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