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Nimfa-mama [501]
3 years ago
12

calculate the acceleration of a 41-kg crate of baseball gear when pulled sideways with a net force of 1255 N. neglect frictional

forces
Physics
1 answer:
lara31 [8.8K]3 years ago
7 0

In his celebrated Second Law of Motion, Newton wrote:

             Net force  =  (mass) x (acceleration).

By the process of "plugging in numbers", we can write

             1255 N  =  (41 kg) x (acceleration)

Now, after dividing each side by (41 kg), we have

             (1255 N) / (41 kg)  =  acceleration.

But  (1255N)/(41kg) = 30.61 m/s² .

So unless we have carelessly blooped the calculations somewhere,
that  30.61 m/s²  is the answer we're looking for.
  
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A missile is fired from a jet flying horizontally at Mach 1 (1100 ft/s). The missile has a horizontal acceleration of 1000 ft/s2
nadezda [96]
Answer: 11,100 ft/s^2

1) Constant acceleration=> uniformly accelerated motion.

2)  Formula for uniformly accelerated motion:

Vf = Vo + at

3) Data:

Vo = 1,100 ft/s
a = 1,000 ft/s^2
t = 10.0 s

4) Solution:

Vf = 1,100 ft/s + 1,000 ft/s^2 * 10.0 s = 1,100 ft/s + 10,000 ft/s

Vf = 11,100 ft/s

6 0
3 years ago
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The force on the spring is F0 and it stores elastic potential energy PEs0. If the spring displacement is tripled to 3x0, determi
Dimas [21]

Answer:

Explanation:

Let initial extension in the spring= x₀

Force on the spring = F₀

Let spring constant = k

Fo = k x₀

Fn = 3k x₀

Fn /Fo = 3

PEs0 ( ORIGINAL)  =1/2 k x₀²

PEsn ( NEW)  =1/2 k (3x₀)²

PEsn / PEs0 = 9

7 0
2 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
What happens to a sample of matter when it’s particles lose kinetic energy?
Mrrafil [7]
What will happen if the sample is the
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5 0
2 years ago
In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo
irinina [24]

Answer:

The torque about the origin is 2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}

Explanation:

Torque \overrightarrow{\tau} is the cross  product between force \overrightarrow{F} and vector position \overrightarrow{r} respect a fixed point (in our case the origin):

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]

\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})

\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})

\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}

4 0
2 years ago
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