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4 years ago

Can velocity and acceleration point in opposite direction?

Physics

2 answers:

ICE Princess25 [194]4 years ago

6 0

Yes, therefore the object would then slow down

neonofarm [45]4 years ago

5 0

Of course !
That's what's happening whenever you're slowing down.

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A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on Mars?

zhannawk [14.2K]

Field strength =force/unit mass
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8 0

3 years ago

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10. Most reabsorption occurs in the

Airida [17]

The answer was B!!:)).

3 0

4 years ago

If an object is thrown upward with an initial velocity of 128 ft/second, then its height after t seconds is given by the follow

IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

$h = 128t-32t^2$ ....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

$\dfrac{dh}{dt}=0$

From equation (I)

$\dfrac{dh}{dt}=128-64t$

$128-64t=0$

$t=\dfrac{128}{64}$

$t=2\ sec$

Now, for maximum height

Put the value of t in equation (I)

$h =128\times2-32\times4$

$h=128\ ft$

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

$128t-32t^2=0$

$t(128-32t)=0$

$128=32t$

$t=4\ sec$

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0

3 years ago

The block of weight 113 lb is pushed with a force of P = 31 lb on top of two cylindrical rollers with a weight of 42 lb and a ra

Lilit [14]

Answer:

The block's speed will be of 5,1 ft/s

Explanation:

For this problem solve it is used the work and energy theorem which state:

$W=dK$

Where W is the net work done and dK the change in the kinetic energy between in the time after and before the force is applied. The net work can expressed as:

$W=F.d$

Here F and d are applied force and the displacement. While the kinetc energy for the i body:

$Ki=1/2*m_{1}V_{1}^2+1/2*I_{1}w_{1}^2$

Where mi,Ii Vi, wi are the mass, the moement of inertia, the linear and the angular velocity for the i body respectively. Therefore to calcule the total change in kinetic energy we take into account all the bodies involves, in this case the block and the two rollers. Assuming fixed rollers but can rotate, and the block with linear movement, the final kinect energy is:

$Kf=1/2*m_{b}V_{b}^2+1/2*I_{r1}w_{r1}^2+1/2*I_{r2}w_{r2}^2$

Note since at the beginning the bodies system is at rest hence the initial kinetic energy is null. In the above equation the subscripts b, r1 and r2 relate to the block and the two rollers.

For solid roller the moment of inertia can be calculated as:

$I_{r}=1/2*m_{r}R_{r}$

Where mr and Rr are the mass and radius rollers. Replacing the moment of inertia in the kinetic energy:

$Kf=1/2*m_{b}V_{b}^2+1/4*m_{r1}R_{r1}w_{r1}^2+1/4*m_{r2}R_{r2}w_{r2}^2$

As at the point of contact there is no displacement between the surfaces of the block and the rollers, in that point the linear velocities of this bodies are the same. So can expressed the linear velocity of the block as function of the rotating speed and the rollers radius:

$V_{b}=w_{r}R$

Entering latest in the new kinetic energy expression:

$Kf=1/2*m_{b}.(w_{r}.R)^2+1/4*m_{r}Rw_{r}^2+1/4*m_{r}Rw_{r}^2$

By doing this the kinetic energy expressed only as function of the roller rotating speed. As the roller rotatings speed are the same, and the masses and the radius are known, the the kinetic energy can expressed as:

$Kf=1/2*113\ lb.\1.57\ ft\ w^2+1/4*42\ lb.\ 1.57\ ft\ w^2+1/4*42\ lb.\ 1.57\ ft\ w^2$