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r-ruslan [8.4K]
3 years ago
12

The force sensor measures the force on the sensor due to the bumper, but the cart's momentum change arises from the force on the

cart due to the bumper. Which of the following facts are needed to assert that the magnitude of these two forces are nearly equal at all times. 1. Their magnitudes differ by the magnitude of the net force on the bumper. 2. The force of friction is small. 3. The net force on the bumper is small. 4. The cart's weight is canceled by the the normal force exerted by the track.
Physics
1 answer:
Irina-Kira [14]3 years ago
5 0

Answer:

Answered

Explanation:

1 and 3 are necessary

Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.

If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.

If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.

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Una moto accelera da ferma a 80 km/h in 4,0 s. A quale forza di inerzia è sottoposto il pilota che ha massa 75 kg?
scoray [572]

A=65 divideded by 55 plus 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Explanation:

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A race car starting from rest accelerates uniformly at a rate of 4.90 meters per squared. What is the cars speed after it has tr
shusha [124]
Ok, we need to find a relation for the speed as it relates to the acceleration.  This is given by the integral of acceleration:

v= \int\limits^{t}_{0} {a} \, dt' =at

Where we have the initial velocity is 0m/s and a will be 4.90m/s².

But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating!  We need to calculate this time by now finding the position function as a function of time.  This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity. 
Position is just the integral of velocity:

s= \int\limits^{}_{} {at} \, dt = \frac{1}{2}at^2

Where the initial velocity and initial position are both zero.

Now we set this position function equal to 200m and find the time, t, it took to get there

\frac{1}{2}(a \frac{m}{s^2} )t^2=200m \\  \\ \frac{1}{2}4.90 \frac{m}{s^2} t^2=200m \\  \\ t^2= \frac{400m}{4.90 \frac{m}{s^2}}=81.63s^2 \\  \\ t= \sqrt{81.63s^2 }  =9.04s

Now let's put t=9.04s into our velocity equation:

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8 0
3 years ago
The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d
r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let F_1\ and\ F_2 are two forces. The resultant of two forces acting on the same point is given by :

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}

Where \theta is the angle between two forces

When \theta=0 i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}

When \theta=90^{\circ} i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}

F_R=\sqrt{F_1^2+F_2^2}

When \theta=180^{\circ} i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}

F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

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The block of wood is 3cm on each side so it is a cube. The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3. density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3
8 0
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