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Bond [772]
3 years ago
13

Is the normal force equivalent to the weight of an object?

Physics
1 answer:
Greeley [361]3 years ago
4 0

Answer:

No.

Explanation:

If there are no applied forces, normal force is usually equivalent to the weight of the object but if there are outside force (force that makes the object to move) especially if it's inclined, then the inclined force would then affect the normal force.

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Find the change in kinetic energy of a 1.0 kg fish leaping to the right at 12.0 m/s.
sp2606 [1]

Answer:

6J

Explanation:

Given parameters:

Mass of fish  = 1kg

Velocity  = 12m/s

Unknown:

Change in kinetic energy = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically given as:

       K.E  = \frac{1}{2}  m v²  

Now, insert the parameters and solve;

  K.E  =  \frac{1}{2}  x 1 x 12  = 6J

The change in kinetic energy is 6J

4 0
3 years ago
Read 2 more answers
Which of these belongs to a tropic level with the most energy?
MrRa [10]

Answer:

Berries is the correct answer because it is the produce in your pyramid and as each living thing is devoured by another there is less energy. For instance the berry has the most energy because it’s energy has just come from the sun. But then an insect eats it and consumes most of its energy but some energy is released into the atmosphere. Then a rodent eats the bug and consumes its energy but yet again some energy is released into the atmosphere. So each time there is less and less energy. Does that help any?

Explanation:

it’s energy has just come from the sun. But then an insect eats it and consumes most of its energy but some energy is released into the atmosphere.

8 0
3 years ago
What is the rate of change of an object’s velocity per unit of time known as?
irga5000 [103]
The rate of change of object's velocity per unit of time is known as acceleration.

Hope this helps you.
6 0
3 years ago
Read 2 more answers
How do you calculate the net force, i need a full explanation PLEASE
Lina20 [59]

Answer:

Once you have drawn the free-body diagram, you can use vector addition to find the net force acting on the object. We will consider three cases as we explore this idea:

Case 1: All forces lie on the same line.

If all of the forces lie on the same line (pointing left and right only, or up and down only, for example), determining the net force is as straightforward as adding the magnitudes of the forces in the positive direction, and subtracting off the magnitudes of the forces in the negative direction. (If two forces are equal and opposite, as is the case with the book resting on the table, the net force = 0)

Example: Consider a 1-kg ball falling due to gravity, experiencing an air resistance force of 5 N. There is a downward force on it due to gravity of 1 kg × 9.8 m/s2 = 9.8 N, and an upward force of 5 N. If we use the convention that up is positive, then the net force is 5 N - 9.8 N = -4.8 N, indicating a net force of 4.8 N in the downward direction.

Case 2: All forces lie on perpendicular axes and add to 0 along one axis.

In this case, due to forces adding to 0 in one direction, we only need to focus on the perpendicular direction when determining the net force. (Though knowledge that the forces in the first direction add to 0 can sometimes give us information about the forces in the perpendicular direction, such as when determining frictional forces in terms of the normal force magnitude.)

Example: A 0.25-kg toy car is pushed across the floor with a 3-N force acting to the right. A 2-N force of friction acts to oppose this motion. Note that gravity also acts downward on this car with a force of 0.25 kg × 9.8 m/s2= 2.45 N, and a normal force acts upward, also with 2.45 N. (How do we know this? Because there is no change in motion in the vertical direction as the car is pushed across the floor, hence the net force in the vertical direction must be 0.) This makes everything simplify to the one-dimensional case because the only forces that don’t cancel out are all along one direction. The net force on the car is then 3 N - 2 N = 1 N to the right.

Case 3: All forces are not confined to a line and do not lie on perpendicular axes.

If we know what direction the acceleration will be in, we will choose a coordinate system where that direction lies on the positive x-axis or the positive y-axis. From there, we break each force vector into x- and y-components. Since motion in one direction is constant, the sum of the forces in that direction must be 0. The forces in the other direction are then the only contributors to the net force and this case has reduced to Case 2.

If we do not know what direction the acceleration will be in, we can choose any Cartesian coordinate system, though it is usually most convenient to choose one in which one or more of the forces lie on an axis. Break each force vector into x- and y-components. Determine the net force in the x direction and the net force in the y direction separately. The result gives the x- and y-coordinates of the net force.

Example: A 0.25-kg car rolls without friction down a 30-degree incline due to gravity.

We will use a coordinate system aligned with the ramp as shown. The free-body diagram consists of gravity acting straight down and the normal force acting perpendicular to the surface.

We must break the gravitational force in to x- and y-components, which gives:

F_{gx} = F_g\sin(\theta)\\ F_{gy} = F_g\cos(\theta)F

gx

​

=F

g

​

sin(θ)

F

gy

​

=F

g

​

cos(θ)

Since motion in the y direction is constant, we know that the net force in the y direction must be 0:

F_N - F_{gy} = 0F

N

​

−F

gy

​

=0

(Note: This equation allows us to determine the magnitude of the normal force.)

In the x direction, the only force is Fgx, hence:

F_{net} = F_{gx} = F_g\sin(\theta) = mg\sin(\theta) = 0.25\times9.8\times\sin(30) = 1.23 \text{ N}F

net

​

=F

gx

​

=F

g

​

sin(θ)=mgsin(θ)=0.25×9.8×sin(30)=1.23 N

7 0
3 years ago
A force is always needed to slow an object down.<br><br> A True<br> B false
Fiesta28 [93]

The answer is TRUE, I'm pretty sure.

5 0
3 years ago
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