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3 years ago

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If an object is resting on a surface it "crunches" the atoms underneath it. According to Newton's Third Law the surface pushes b

ack on the object. What is the name given to this force that acts perpendicular to the surface?

1 answer:

alex41 [277]3 years ago

7 0

Answer:

Normal force

Explanation:

Normal force is the force that surfaces employ so that solid objects are disallowed from passing through each other.

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A hot-air balloon is floating above a straight road. To estimate their height above the ground, the balloonists simultaneously m

ki77a [65]

We know that
tan(∅) = y/x; where y will be the height of the balloon and x will be the distance from the milestone marker.
The marker that is further away will produce a smaller angle of depression.
tan(18) = height / x
x = height / tan(18)

The second marker is x + 1 miles away:
tan(16) = height / x + 1; substituting x:
tan(16) = height / (height/tan(18) + 1)
tan(16) x [(height / tan(18) + 1] = height
0.88h + 0.29 = h
h = 2.42 miles

8 0

3 years ago

POSSIBLE POINTS: 1

faust18 [17]

Answer:

Pushes and pulls refer to the force that attracts or repels certain other materials without actually touching them.

Explanation:

Pushes and pulls are the forces exerted by the magnet on certain materials around it without, actually touching them. This push and pull is exerted through a region around the magnet called its magnetic field. The strength of this push and pull force is determined by, the strength of the magnetic field. A strong push or pull force is exerted by a strong magnetic field, and in turn a strong magnet and, a weak push and pull force is exerted by a weak magnetic field and, in turn a weak magnet. A push force is a repulsion while a pull force is an attraction. When a magnetic object is in the region of the magnetic field, it either attracts or is repelled away from the source of the magnetic field.

7 0

3 years ago

Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about t

Ronch [10]

1. The bowling ball have more potential energy as it sit on top of the building

2.The bowling ball have same potential energy and kinetic energy as it is half way through its fall

3. The bowling ball have more kinetic energy just before it hits the ground

4. The potential energy of the bowling ball as it sits on top of the building is 784J

5. The potential energy of the ball as it is half way through the fall, 20 meters high is 392J

6. The kinetic energy of the ball as it is half way through the fall is 392J

7. The kinetic energy of the ball just before it hits the ground is 784J

Explanation:

Calculating potential energy and kinetic energy for all the instances,

1. ball on top of a 40 meters tall building

Potential energy at the top of building with a height of 40m = mgh

P.E = mgh =2*9.8*40= 784J

At the top pf the building since v=0 kinetic energy is zero

2. half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second

Potential energy when it is half way through fall = mgH

where H represents new height that is equal to 20m

hence P.E=mgH=2*9.8*20= 392J

Kinetic energy of the ball is $\frac{1}{2} mv^{2} = \frac{1}{2} *2*19.8^{2}$ =392.04J

3. Just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.

The potential energy of ball just before it hits the ground = mgh= 2*9.8*0=0J

kinetic energy = $\frac{1}{2} mv^{2}$ = 784J

3 0

3 years ago

1. A Passenger car (m= 10.0 kg) and a flat car (m= 7.5 kg) collide and move off as

grin007 [14]

Answer:

use a calculator to solve that

6 0

3 years ago

A ball is rolled of the edge of a table with a

tatiyna

Answer:

Approximately $10\; \rm m \cdot s^{-1}$ at $5.6^\circ$ below the horizon.

Horizontal component of velocity: $10\; \rm m \cdot s^{-1}$ .
Vertical component of velocity: $0.981\; \rm m\cdot s^{-1}$ (downwards.)

(Assumption: air resistance on the ball is negligible; $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the air resistance on the ball is negligible. The horizontal component of the velocity of the ball would stay the same at $10\; \rm m \cdot s^{-1}$ until the ball reaches the ground.

On the other hand, the vertical component of the ball would increase (downwards) at a rate of $g = 9.81\; \rm m\cdot s^{-2}$ (where $g$ is the acceleration due to gravity.) In $0.1\; \rm s$ , the vertical component of the velocity of this ball would have increased by $9.81\; \rm m \cdot s^{-2} \times 0.10\; \rm s = 0.981\; \rm m \cdot s^{-1}$ .

However, right after the ball rolled off the edge of the table, the vertical component of the velocity of this ball was $0\; \rm m\cdot s^{-1}$ . Hence, $0.10\; \rm s$ after the ball rolled off the table, the vertical component of the velocity of this ball would be $0\; \rm m \cdot s^{-1} + 0.981\; \rm m\cdot s^{-1} = 0.981\; \rm m \cdot s^{-1}$ .

Calculate the magnitude of the velocity of this ball. Let $v_{x}$ and $v_{y}$ and denote the horizontal and vertical component of the velocity of this ball, respectively. The magnitude of the velocity of this ball would be $\displaystyle \sqrt{{v_x}^{2} + {v_y}^{2}}$ .

At $0.10\; \rm s$ after the ball rolled off the table, $v_x = 10\; \rm m \cdot s^{-1}$ while $v_y = 0.981\; \rm m \cdot s^{-1}$ . Calculate the magnitude of the velocity of the ball at this moment:

$\begin{aligned} \| v \| &= \sqrt{{v_x}^{2} + {v_y}^{2}} \\ &= \sqrt{\left(10\; \rm m \cdot s^{-1}\right)^{2} + \left(0.981\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 10.0\; \rm m\cdot s^{-1}\end{aligned}$ .

Calculate the angle between the horizon and the velocity of the ball (a vector) at that moment. Let $\theta$ denote that angle.

$\displaystyle \tan \theta = \frac{\text{rise}}{\text{run}}$ .

For the vector representing the velocity of this ball:

$\displaystyle \tan \theta = \frac{0.981\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-1}} = 0.0981$ .

Calculate the size of this angle:

$\theta = \arctan 0.0981 \approx 5.62^\circ$ .

Notice that the vertical component of the velocity of this ball at that moment points downwards (towards the ground.) Hence, the corresponding velocity should point below the horizon.

5 0

3 years ago