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3 years ago

Which of the following is a contact force?

Physics

1 answer:

3241004551 [841]3 years ago

3 0

Answer:

b friction

Explanation:

Contact forces

Contact forces are forces that act between two objects that are physically touching each other. Examples of contact forces include:

Reaction force

An object at rest on a surface experiences reaction force. For example, a book on a table.

A box rests on a table. There are two arrows, equal in size but going in opposite directions, up and down, from the point where the box meets the table.

Tension

An object that is being stretched experiences a tension force. For example, a cable holding a ceiling lamp.

A box hangs from a rope. Two arrows which are equal in size act upwards and dowards from the top and bottom of the rope.

Friction

Two objects sliding past each other experience friction forces. For example, a box sliding down a slope.

A box rests on an incline. There are three arrows; one acting vertically downwards from the centre of the box’s base. One arrow acts perpendicular to the incline. One arrow acts up the incline.

Air resistance

An object moving through the air experiences air resistance. For example, a skydiver falling through the air.

A box falls from the sky. Two arrows, equal in size and opposite in direction act upwards from the box and downwards from the box

When a contact force acts between two objects, both objects experience the same size force, but in opposite directions. This is Newton's Third Law of Motion.

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Some jovian planets give off more energy than they receive because ofa. fusion in their coresb. tidal heatingc. ongoing contract

faltersainse [42]

Answer:

the energy emission process is due to nuclear fusion

Explanation:

In the planetary system the energy emitted by bodies is a part of the energy received by a hot body (star), this star is what creates the entire energy cycle of a given planet. This creates the entire tidal movement, continental drift

If the planets reach a certain size the gravitational force near their center is so great that the atoms are crushed and some beginnings of fusion processes are formed where some atoms are transformed from hydrogen to helium releasing a certain amount of energy and heat that believe that the planet emits more energy than it receives from its star.

In summary, the energy emission process is due to nuclear fusion

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3 years ago

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If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k

loris [4]

Recall the definition of the cross product with respect to the unit vectors:

i × i = j × j = k × k = 0

i × j = k

j × k = i

k × i = j

and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)

Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have

a × b = (8i + j - 2k) × (5i - 3j + k)

a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)

… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)

… … … … + 8 (i × k) + (j × k) - 2 (k × k)

a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)

a × b = - 5k - 10j - 24k - 6i - 8j + i

a × b = -5i - 18j - 29k

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3 years ago

A mule is harnessed to a sled having a mass of 246 kg, including supplies. The mule must exert a force exceeding 1210 N at an an

zimovet [89]

(a) 1800 N

The equation of the forces along the vertical direction is:

$F sin \theta + N - mg = 0$

where

$F sin \theta$ is the component of the applied force along the vertical direction

N is the normal force on the sled

mg is the weight of the sled

Substituting:

F = 1210 N

m = 246 kg

$\theta = 30.3^{\circ}$

We find N:

$N=mg-F sin \theta = (246)(9.8)-(1210)(sin 30.3^{\circ})=1800 N$

(b) 0.580

The equation of the forces along the horizontal direction is:

$F cos \theta - \mu_s N = 0$

where

$F cos \theta$ is the horizontal component of the push applied by the mule

$\mu_s N$ is the static frictional force

Substituting:

F = 1210 N

N = 1800 N

$\theta = 30.3^{\circ}$

We find $\mu_s$ , the coefficient of static friction:

$\mu_s = \frac{F cos \theta}{N}=\frac{(1210)(cos 30.3^{\circ})}{1800}=0.580$

(c) 522 N

In this case, the force exerted by the mule is

$F= 6.05 \cdot 10^2 N = 605 N$

So now the equation of the forces along the horizontal direction can be written as

$F cos \theta - F_f = 0$

where

$\theta=30.3^{\circ}$

and $F_f$ is the new frictional force, which is different from part (b) (because the value of the force of friction ranges from zero to the maximum value $\mu N$ , depending on how much force is applied in the opposite direction)

Solving the equation,

$F_f = F cos \theta = (605)(cos 30.3^{\circ})=522 N$

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3 years ago

Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.

Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

$distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m$

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

$displacement = +1.0 m-0 m=+1 m$

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3 years ago

The unit for work can be written as __________ or as the nickname ______

Arada [10]

N or joule cuz joule is newton’s

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3 years ago

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