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3 years ago

Convert 75.0 degrees Celsius into Fahrenheit

Physics

1 answer:

Elis [28]3 years ago

6 0

Answer: (75°C × 9/5) + 32 = 167°F

Explanation: Thats the formula

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Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?

Yakvenalex [24]

None because Neon is already stable

8 0

3 years ago

A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?

Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration = 24m/s²

Unknown:

Force needed by the cheetah to run = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

Force = mass x acceleration

Insert the given parameters and solve;

Force = 12 x 24 = 288N

7 0

3 years ago

If a neutral material like fur is rubbed on another neutral object like a rubber ball what is the resulting charges of the objec

alexdok [17]

Answer:

Explanation:

they both have a neutral charge so they couldn't be positive or negative since that wouldn't come from anywhere

8 0

2 years ago

How long does it take to travel a distance of 672km at a speed of 95km/h?

Brilliant_brown [7]

Answer:

7.07 hours

Explanation:

divide the distance by the speed

so in this case, divide 672 by 95

6 0

3 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