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2 years ago

A 2.0-kilogram ball rolls down a ramp. if the ball accelerates at a rate of 12 m/s2, the net force causing the acceleration is:

Physics

1 answer:

ehidna [41]2 years ago

7 0

Answer:

Explanation:

f = ma

2 x 12 = 24

answer could differ since it's rolling down a ramp. if an angle is given our approach differs.

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A radioactive isotope of the element potassium decays to produce argon. If the ratio of argon to potassium is found to be 31:1,

iris [78.8K]

Answer:

Explanation:

Argon to potassium ratio after 1 half life = 1:1

After 2 half lives = 75/25= 3:1

After 3 half lives = 87.5/12.5= 7:1

After 4 half lives = 93.75/6.25 = 15:1

After 5 half lives = 96.875/3.125 = 31/1

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3 years ago

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Andrews [41]

Answer:

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Explanation:

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3 years ago

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Why did the potassium permanganate crystals start to dissolve in water without being stirred or shaken

KonstantinChe [14]

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3 years ago

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A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn

olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

$v_{f} = v_{i} + a*t\\$

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0

4 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago