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3 years ago

What kind of quantity is displacement?

Physics

1 answer:

Serhud [2]3 years ago

8 0

Answer:

calar quantity, length of path. displacement: vector quanity, "as the crow flies" difference between start and finish regardless of path taken. Term.

Explanation:

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What is the force that opposes motion and works against the downward pull? A) friction B) gravity C) weight D) acceleration

charle [14.2K]

I'm not entirely sure, but I believe it is A Friction. because gravity pulls down, weight isn't a force, and acceleration doesn't oppose motion

6 0

4 years ago

Which of the following describes an electric conductor?

IrinaVladis [17]

Yo sup??

the correct answer is option C ie

a material that has a low resistance and allows charges to move freely

this is a basic property shown by conductors

Hope this helps

6 0

3 years ago

Read 2 more answers

a block is pushed up a frictionless 40 incline. if the initial velocity after the push is 5.00 m/s. how far along the incline do

Lana71 [14]

Answer:

d=1.982m, t=1.019s

Explanation:

There are different approaches we can take to solve this problem. You could either solve this by using conservation of energy or by taking a kinematic approach. I'll solve this by using kinematics. So, the very first thing we need to do in order to solve this is do a drawing of the situation so we can analyze it better. (See attached picture).

So, since we are talking about an inclined plane, we can see that the force of gravity is being split into an x and y components if we incline the axis of coordinates. Taking this into account we can see that:

$\sum F_{x}=ma_{x}$

Since there is no friction in our system, then the only force acting upon the box is the force of gravity, or weight. Since we are taking the upwards direction as the positive direction of movement, we can say that the force of gravity is excerting a negative influence on our box, so this acceleration will be negative, so our sum of forces will now look like this:

$-mg sin(40^{o})=ma$

we can cancel the masses out so we can see that:

a=-g sin(40°)

$a=-9.81m/s^{2} sin(40^{o})$

We have now enough information to solve our problem.

we can take the following equation to find the distance the block travels up the incline:

$x=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

we know the final velocity must be zero, so we can use the provided data to solve our formula:

$x=\frac{(0)^{2}-(5m/s)^{2}}{2(-9.81m/s^{2})sin 40^{o}}$

which yields:

x=1.982m

In order to find the time it takes for the block to return to its original position we can use the following formula:

$x=V_{0}t+\frac{1}{2}at^{2}$

since x=0 is the starting point we can use that to solve our equation:

$0=5t+\frac{1}{2}(-9.81sin 40^{o})t^{2}$

which simplifies to:

$0=5t-4.905t^{2}$

which can now be solved for t

t(5-4.905t)=0

t=0 and 5-4.905t=0

t=0 and $t=\frac{5}{4.905}=1.019$

so the time it takes the block to return to its original position is

t= 1.019s

6 0

3 years ago

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t

Tcecarenko [31]

Answer:

The amount of heat transfer is 21,000J .

Explanation:

The equation form of thermodynamics is,

ΔQ=ΔU+W

Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.

Substitute 0 J for W and 0 J for ΔU

ΔQ = 0J+0J

ΔQ = 0J

The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change

The heat transfer is,

ΔQ=Q (in ) −Q (out )

Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)

ΔQ=(19000J+2000J)−(0J)

=21,000J

Thus, the amount of heat transfer is 21,000J .

8 0

3 years ago

Read 2 more answers

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$