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2 years ago

How can a karate expert break a concrete block

Physics

1 answer:

FinnZ [79.3K]2 years ago

8 0

Many ways, but some of the most famous are kicks (side, back, front, snap) or a smash.

Hope it helped! :)

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A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h

solniwko [45]

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

Distance of ship in west from harbor = ?

now,

Distance of the ship in the west direction

D = d cos θ

d = 575 Km

θ = 23.5°

inserting all the values

D = 575 x cos 23.5°

D = 575 x 0.91706

D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km

6 0

2 years ago

A draft is made of a plastic block with a density of 650kg/m^3, and its dimensions are 2.0mx3.0mx5.0m ,what is the raft's appare

mixas84 [53]

Weight in water = mass of block - mass of volume of displaced water

8 0

2 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

3 years ago

A stiff wire 32.5 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l

Flauer [41]

Answer: i personally think it would be 789

Explanation:

Hopefully I am right

4 0

3 years ago

The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi

adoni [48]

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.

5 0

2 years ago