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77julia77 [94]
2 years ago
8

A truck initially traveling at a speed of 22 meters per second increases speed at a constant rate of 2.4 meters per second^2 for

3.2 seconds. what is the total distance traveled by the truck during this 3.2 second time interval
Physics
1 answer:
Usimov [2.4K]2 years ago
8 0
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:

d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m 
Hope the answer helps. 
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Answer:128 N

Explanation:

Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .

If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.        

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You drop a 6.0x10^-2 kg ball from height of 1.0m above hard flat surface. Ball strikes surface and energy decreases by 0.14J, th
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2 years ago
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Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

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Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

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8 0
2 years ago
A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

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r=(0,1)

angle = 90°

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b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

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the same as e):

L = 5kgm/s

g)

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L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

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the same as g):

L = 5kgm/s

hope this helps!!

4 0
2 years ago
A What CE describes the way energy is stored in a sandwich​
SVETLANKA909090 [29]
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