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Liula [17]
3 years ago
10

In unit-vector notation, what is the net torque about the origin on a flea located at coordinates (-2.0, 4.0 m, -1.0 m) when for

ces F1 = (-4.0 N) k and F2 = (-5.0 N) j act on the flea?
Physics
1 answer:
kifflom [539]3 years ago
6 0

Answer:

Torque, \tau=-21i-8j+10k

Explanation:

Given that,

Coordinates, r=(-2i+4j-k)\ m

Force acting on the flea, F=(0i-5j-4k)\ N

Let \tau is the net torque about the origin on a flea. It is given by the following formula :

\tau=r\times F

\tau=(-2i+4j-k)\times (0i-5j-4k)

\tau=\begin{pmatrix}-21&-8&10\end{pmatrix}

\tau=-21i-8j+10k

So, the net torque about the origin on a flea is -21i-8j+10k. Hence, this is the required solution.

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Answer:

50m

   

Explanation:

Given parameters:

Initial velocity  = 20m/s

Acceleration  = 4m/s²

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Unknown:

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Solution:

To solve this problem use the expression below;

   

     v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

  final velocity = 0

 Insert the parameters and solve;

  0²  = 20² + 2 x 4 x s

   -400  = 8s

         s  = 50m

   Disregard the negative sign because distance cannot be negative.

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The temperature of a 5.0 kg block increases by 3 degrees C when 2,000 J of thermal energy are added to the block. What is the sp
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4 years ago
A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​
just olya [345]

Answer:

5 m/s

Explanation:

Given that,

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​.

It is assumed to find the resultant velocity of the vehicle. Let east side is positive and west is negative. So,

v=v_1+v_2\\\\=20+(-15)\\\\=5\ m/s

Hence, the resultant velocity of the vehicle is equal to 5 m/s.

6 0
3 years ago
Which example best illustrates kinetic energy?
mihalych1998 [28]

Answer:

A.

Explanation:

In all actuality. The car may not be fully moving but since it's on a hill and the earth moves it makes the most sense. Also since the car is at a slope by the law of motion what goes up must come down so the car is in fact moving.

8 0
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A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio
lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

W = F\times d

KE = 0.5\times m\times v^2

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

W = F \times d = 2.62 N \times 100 m

W = 261.6 N\times m

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

0.5 m\times v2^2 = 0.5 m\ v1^2 - W

Now solve for v2

v2 = \sqrt{v1^2 - {\frac{2W}{M}}}

= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

Ff = Us\times N

Now solve for Us

= \frac{Ff}{N}

= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}

= 0.00494

4 0
3 years ago
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