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2 years ago

15

A brick is lying on a table in a state of static equilibrium. If the mass of the brick is 7.52 kilograms, what is the normal for

ce exerted by the table on the brick?

1 answer:

WARRIOR [948]2 years ago

7 0

Answer:

-mg = -7.52*9.8 = - 73.696N

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A freight car moves along a friction less level railroad track at constant speed. The car is open on top. A large load of coal i

maxonik [38]

Answer:

The velocity of the freight car decreases.

Explanation:

This question is answered by the conservation of momentum principle.

When the freight car is moving at a certain speed, it has a constant momentum.

We will call this M1.

The equation for M1 will be:

M1 = Mass * Speed

Now when the coal is dumped into the freight car, the Mass increases.

Since conservation of momentum states that the momentum will remain the same. We have:

M1 = (Mass of freight + Mass of coal) * Speed

Since M1 is constant, if the mass increases, the speed had to decrease to keep the equation true.

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3 years ago

In a electrically natural atom of any element there are equal numbers of

Usimov [2.4K]

Answer: Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion.

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3 years ago

HELP ASAP!! i’ll mark you the brainliest!!

Triss [41]

Answer:

yes

Explanation:

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2 years ago

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

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3 years ago

During the flight of the ball, what is the direction of its Acceleration due to gravity

vekshin1

Regardless of what direction an object is moving, the acceleration
due to gravity is always directed toward the center of the Earth.
That's the direction commonly known as "down".

4 0

3 years ago