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3 years ago

Lực hút trái đất là dì

Physics

1 answer:

nataly862011 [7]3 years ago

5 0

Lực hút trái đất gọi Là trọng lực

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PLEASE HELP! WILL GIVE BRAINLY Which of the four fundamental forces acts over very long distances but does

Sergeu [11.5K]

Answer:

Explanation:

just did it

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3 years ago

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How much time does it take a power drill accelerating at 64.3 rad/s2 to achieve

Diano4ka-milaya [45]

Answer;

= 0.244 seconds

Explanation;

500 rpm is equivalent to; 1500 × 2π radians per minute

= 9424.8 rad/minute

To get revolutions per second we divide by 60

= 9424.8/60

= 157.08 radians per second.

Then we divide by 64.3 rad/s^2 to get time;

= 157.08/64.3

= 0.244 seconds.

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3 years ago

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what is the equation for newton’s second law of motion, in terms of mass, velocity and time, when the mass of the system is cons

Sergio039 [100]

Answer:

Newton's second law of motion is F = ma, or force is equal to mass times acceleration.

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3 years ago

A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction betw

umka2103 [35]

Answer:

Explanation:

Using the Newtons law of motion;

$\sum Fx = ma_x\\Fm - Ff = ma_x$

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

$Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69$

Hence the coefficient of kinetic friction between the box and floor is 0.69

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3 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

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3 years ago