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kvasek [131]
3 years ago
14

There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

l block is 25 kg, and the large block is 50 kg. The blocks are to the left of a hill.The hill is 10 meters tall. (Drawing not to scale)a) The small block collides with the large block and sticks together. How fast must the initial velocityof the small block be so that the two blocks just reach the top of the hill. (Assume no friction).b) What was the total impulse the small block exerted on the large block in order to get it going fastenough
Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2  x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75  x 42 / 3

= 1050 Ns.

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0.80865 Hz

1.23662 seconds

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o
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Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges q_1 and q_2 separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{r^2}

Where k is the proportional constant of value

k=9*10^9\ N.m^2/c^2

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

\displaystyle F_1=\frac{k\ Q\ Q}{d^2}

\displaystyle F_1=\frac{k\ Q^2}{d^2}

The force between charges separated 2d is

\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}

\displaystyle F_2=\frac{k\ Q^2}{4d^2}

And the force between the charges A and D is

\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}

\displaystyle F_3=\frac{k\ Q^2}{9d^2}

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_A=-\frac{41}{36}F_1

Charge B. It receives force to the right from A and D and to the left from C

\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}

\displaystyle F_B=\frac{1}{4}F_1

Charge C. It receives forces to the right from all charges.

\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}

\displaystyle F_C=\frac{9}{4}F_1

Charge D. It receives forces to the left from all charges

\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}

\displaystyle F_D=-\frac{49}{36}F_1

Comparing the magnitudes of each force is just a matter of computing the fractions

\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36

a.

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

b.

The ratio of the greatest to the smallest net force is

\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9

The greatest force is 9 times the smallest net force

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3 years ago
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