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ladessa [460]
3 years ago
6

What is newtons first law? what is the other name for it?

Physics
1 answer:
Leno4ka [110]3 years ago
5 0

Answer:

Explanation:

An object will remain at rest or in uniform in a straight line upon an external force

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Answer:

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Explanation:

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4 years ago
If a ball is dropped from a height of 10 m, hits the ground and bounces back up to a maximum height of 7.5 m, describe how the E
Greeley [361]

Answer:hmmmm

Explanation:

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3 years ago
A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
Tamiku [17]

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

                                                  F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

4 0
3 years ago
Buhrs atomic model differed from ruthofords because it explained that
Scrat [10]

Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.

6 0
3 years ago
Read 2 more answers
The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris
Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

        \sum F_{x} = 0

        T + F_{AB} Sin 60 = 0 ...... (1)

       \sum F_{y} = 0

       F_{AB} Cos 60 + W = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

               \sigma_{allow} = \frac{T}{A}

       T = (20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}

          = 3925 kip

From equation (1),   F_{AB}Sin (60^{o})  = -3925

               F_{AB} \times -0.304 8 = -3925

             F_{AB} = 12877.29 kip

From equation (2),    -12877.29 (Cos 60) + W = 0

         -12877.29 kip \times \frac{1}{2} + W = 0

                           W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0
3 years ago
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