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sweet [91]
3 years ago
7

A Thomson's gazelle can reach a speed of 13 m/s in 3.0 s. A lion can reach a speed of 9.5 m/s in 1.0 s. A trout can reach a spee

d of 2.8 m/s in 0.12 s. Which equation from the formula sheet should you use to determine which animal has the largest acceleration?
Physics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:

The acceleration of Trout is largest.

Explanation:

Speed of Thomson's gazelle, v = 13 m/s in 3 s. Let us assumed that initial speed is 0. Acceleration is given by :

a_T=\dfrac{v-u}{t}\\\\a_T=\dfrac{v}{t}\\\\a_T=\dfrac{13\ m/s}{3\ s}\\\\a_T=4.34\ m/s^2

Speed of lion is 9.5 m/s in 10 s. Let us assumed that initial speed is 0. Acceleration is given by :

a=\dfrac{v-u}{t}\\\\a_l=\dfrac{v}{t}\\\\a_l=\dfrac{9.5\ m/s}{1\ s}\\\\a_l=9.5\ m/s^2

A trout can reach a speed of 2.8 m/s in 0.12 s. Let us assumed that initial speed is 0. Acceleration is given by :

a=\dfrac{v-u}{t}\\\\a_t=\dfrac{v}{t}\\\\a_t=\dfrac{2.8\ m/s}{0.12\ s}\\\\a_t=23.34\ m/s^2

It is clear that the acceleration of Thomson's gazelle, lion and trout is 4.34\ m/s^2, 9.5\ m/s^2\ and\ 23.34\ m/s^2 respectively. So, the acceleration of Trout is largest.

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What acceleration will you give to a 24.5 kg
Ludmilka [50]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                   \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                             F = Force = 78,3 N

                                            m = Mass = 24,5 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

\boxed{a=78,3\ N / 24,5\ kg}

⇒ Resolving

\boxed{a=3.19\ m/s^{2}}

Result:

The aceleration is <u>3,19 meters per second squared (m/s²)</u>

Good Luck!!

4 0
3 years ago
You are pushing a heavy box across a rough floor. when you are initially pushing the box and it is accelerating, (a) you exert a
kodGreya [7K]

The answer is C. If the box is accelerating, that means that the amount of force you are exerting is greater than the force of the box.

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A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.
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A) 

     It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s
 
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\Delta v=  \frac{\Delta S}{\Delta t}  \\ v_x= \frac{36}{2}  \\ \boxed {v_{x}=18m/s}


B)
 
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v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}
  
     The angle of impact is given by:

cos(\theta) =\frac{v_{x}}{v_{y}}  \\ cos(\theta) = \frac{18}{20}  \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}


If you notice any mistake in my english, please let me know, because i am not native.

7 0
3 years ago
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E = 500 N/C, t = 54 ns = 54 x 10^-9 s,

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For electron:

u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.

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Answer:

30Km/h

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