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3 years ago

Examples of forces in motion that you can see?

Physics

1 answer:

lidiya [134]3 years ago

6 0

Answer:

tensional force

Explanation:

It usually exist in fluid or any liquid substance where by an object is put before it

e.g An ant can walk on water without submersed in it.

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If you placed a dot on a world map to represent the location of every earthquake in the last 100 years, where would you see eart

Vedmedyk [2.9K]

If we are to place dots to teh places that have been struck by an earthquake these past 100 years, the dots would be concentrated in the east and southeast Asia region. This is because of the presence of the Pacific ring of fire. This is a major area in the Pacific Ocean where most of the earthquakes are likely to occur.

6 0

2 years ago

Hell please thanks!!!!!!’

Slav-nsk [51]

Answer:

liquid phase

Explanation:

it is liquid phase because molecules are not that tightly packed as solid and not that far away from each other as in gas phase.

5 0

3 years ago

Where is the distance on a position time graph

S_A_V [24]

Answer:

The distance represents the difference of the first position and last position of the body.

Explanation:

For example, if y axis represents the position axis, and the first position is 3, second 9 we can see that the distance is a (positive) projection of one position into another. 9-3=6

Hope this helps.

4 0

2 years ago

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points

Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

5 0

3 years ago

*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa

Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
This force is not a new force, is just the net force aiming to the center of the circle.
In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
So, we can write the following expression:

$F_{cent} = F_{n} - F_{g} (1)$

It can be showed that the centripetal force is related to the speed by the following expression:
$F_{cent} = m*\frac{v^{2}}{r} (2)$

The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
Replacing (2) in (1), and solving for Fn, we get:

$F_{n} = m*\frac{v^{2} }{r} + m*g (3)$

Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

$F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)$

Replacing Fg by its value, simplifying, and solving for v, we get:

$v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)$

3 0

2 years ago