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VikaD [51]
2 years ago
14

5. Calculate the acceleration of a 2 kg block across a table if you push with a force of 20 N and the frictional force is 4 N.

Physics
1 answer:
frozen [14]2 years ago
8 0

Answer:

Explanation:hey do you go to GOC too?

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1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
dimulka [17.4K]

Answer:

The acceleration of the crate is 0.3356\,\frac{m}{s^2}

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

F=m\,a

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}

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3 years ago
Objects falling through the air experience a type of friction called air resistance. true or false
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Answer:

True

Explanation:

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A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

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If 100 grams of vinegar and 5 grams of baking soda are poured in a container, a small amount of gas will be produced. What will
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