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2 years ago

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Newton’s Second Law establishes the relationship between mass, net applied force, and acceleration given by F=ma. Consider a 4 k

g box of holiday candy on a horizontal surface such as a table. There is a 10N applied force to the right and a 2N force to the left. The box accelerates with a magnitude of _________?

1 answer:

netineya [11]2 years ago

7 0

Answer:

a= 2 m/s^2

Explanation:

take to the right as positive

let Ftot be the total forces acting on the box , m be the mass of the box and a be the acceleration of the box.

Ftot = 10 - 2 = 8 N

and,

Ftot = ma

a = Ftot/m

= 8/4

= 2 m/s^2

therefore, the acceleration of the box is of magnitude of 2 m/s^2.

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Answer:

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Most water vapor enters the atmosphere via evaporation and transpiration. Evaporation occurs when a single water molecule on a liquid water surface gains enough kinetic energy (often by solar radiation) to break the bond which holds the molecules together. Really Hopes this helps!

Explanation:

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2 years ago

Jobisdone [24]

Hey!!

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3 years ago

Read 2 more answers

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

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2 years ago

It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f

goldenfox [79]

Answer:

B) the change in momentum

Explanation:

Impulse is defined as the product between the force exerted on an object (F) and the contact time ( $\Delta t$ )

$I=F \Delta t$

Using Newton's second law (F = ma), we can rewrite the force as product of mass (m) and acceleration (a):

$I=(ma) \Delta t$

However, the acceleration is the ratio between the change in velocity ( $\Delta v$ ) and the contact time ( $\Delta t$ ): $a=\frac{\Delta v}{\Delta t}$ , so the previous equation becomes

$I=m \frac{\Delta v}{\Delta t}\Delta t$

And by simplifying $\Delta t$ ,

$I=m \Delta v$

which corresponds to the change in momentum of the object.

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3 years ago

The uniform 100-kg I-beam is supported initially by its end rollers on the horizontal surface at A and B. by means of the cable

Ann [662]

I attached a free body diagram for a better understanding of this problem.

We start making summation of Moments in A,

$\sum M_A = 0$

$P(6cos\theta)-981(4cos\theta)=0$

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Then we make a summation of Forces in Y,

$\sum F_y = 0$

$654+R-981 = 0$

$R=327N$

At the end we calculate the angle with the sin.

$sin\theta = \frac{3m}{4m+2m+2m} = \frac{3m}{8m}$

$\theta = 22.02\°$

8 0

3 years ago