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3 years ago

A ball rolls to the right across a grassy field slowing down as it does so. Which of the following must be true?

Physics

2 answers:

In-s [12.5K]3 years ago

5 0

Answer: i could prolly answer if it made more sense. which of the following where?

Zepler [3.9K]3 years ago

4 0

Answer:

D: the ball is not accelerating because it is slowing down

Explanation:

if the was no friction, then the ball would most likely pick up speed and never stop, like it would with it being on a hill covered in ice, it would pick up speed and never slow down (well, it would slow down, just at a point where it looks like it would never stop.

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Which of the following correctly describes the relationship between speed and velocity?

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The answer is B

Explanation:

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3 years ago

Read 2 more answers

A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to re

Ilya [14]

Answer:

a) F = -1035.385 N

b) Backwards

c) s = 15.60 m

Explanation:

Given information

$u$ = Initial Speed of Car = 15.0 m/s

$v$ = Final Speed of Car = 9.00 m/s

$t_{b}$ = Breaking Time = 1.30 s

$m$ = Mass of Car = 1040 kg

Part (a)

To find the force exerted on the car we use the following formula

$F = ma$

Where

$F$ = Force = unknown

$m$ = Mass of Car = 1040 kg

$a$ = Acceleration of Car / Deceleration of Car = unknown

To find the force (F) we need to first find the deceleration rate (a)

To find the deceleration rate we use the following formula

$a = \frac{v - u}{t}$

Inputting the given values

$a = \frac{15 - 9}{1.30} \\ a = -4.615$

To find the force

$F = ma \\ F = (1040)(-4.615) \\ F = (1040)(-4.615) \\ F = -1035.385 N$

Part (b)

Since the value of F is negative this means the the force was opposite the direction of motion, hence the force was backwards.

Part (c)

To find the total distance the car moved while braking we use the following formula

$v^2 = u^2 + 2as$

Where

$s$ = distance traveled

Inputting the values given

$(9)^2 = (15)^2 + 2(-4.615)s$

$s = 15.60 m$

6 0

3 years ago

A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles

slamgirl [31]

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035 × v² + 1/2 × 40 × 0.012²

0.032 - 1/2 × 40 × 0.012² = 1/2 × 0.035 × v²

0.02912 = 1/2 × 0.035 × v²

1/2 × 0.035 × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm, v ≈ 1.28996 m/s

E = PE + KE

0.032 = 1/2 × 40 × 0.025² + KE

KE = 0.032 - 1/2 × 40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.

4 0

3 years ago