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Mariulka [41]
3 years ago
7

Can someone check if what i wrote so far makes sense and if i made a mistake.

Physics
2 answers:
Serjik [45]3 years ago
7 0
I don't really know what it's about but everything looks okay to me. There might be some mistakes on the last sentence but i'm not completely sure.
garri49 [273]3 years ago
5 0
It looks great, except, I would rewrite the first sentence as "Everett knew it was dangerous to be within close proximity to a cheetah". Also, in the second sentence, I would get rid of the words "at the moment". They aren't really necessary. That's all. Everything else looks perfect. Good job introducing the subject, by the way.
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Which way will the people on the right move? Why? Plz answer
Scrat [10]
Probably to the right because the force on the left is greater and will therefore overpower the 2N force and push to the right
3 0
3 years ago
An airplane starts at rest and accelerates down the runway for 20 s. At the end of the runway, its velocity is 80 m/s north. Wha
goldenfox [79]
Vi = vf + ( a.t)
 0 m/s (rest) = 80 m/s + (a.20s.-1)

a.20.-1= -80, we check if this is true; = 0 m/s= 80 + -80 = 0 ITS TRUE.

so a.20.-1= -80, a= 80/20 , Answers a= 4 m/s 
5 0
3 years ago
A car starts from 0 m along a road and accelerates at 0.5 m/s^2 to the right. A second car starts from 1000 m along the road and
Brrunno [24]

Answer:

e) 31.6 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow s_1=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_1=\frac{1}{2}0.5t^2\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow s_2=0\times t+\frac{1}{2}\times 0.5\times t^2\\\Rightarrow s_2=\frac{1}{2}1.5t^2\ m

s_1+s_2=1000

\\\Rightarrow 1000=\frac{1}{2}0.5t^2+\frac{1}{2}1.5t^2\\\Rightarrow 1000=\frac{0.5t^2+1.5t^2}{2}\\\Rightarrow 1000=\frac{2t^2}{2}\\\Rightarrow 1000=t^2\\\Rightarrow t=\sqrt{1000}\\\Rightarrow t=31.6\ s

Time taken by the cars to meet 31.6 seconds.

5 0
3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
Goshia [24]

Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

6 0
3 years ago
One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from
Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

s_1 = v_1t = 8t

Similarly the distance traveled by the 2nd skater after t seconds is

s_2 = v_2t = 10t

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

s_2 = s_1 + 80

We can substitute s_1 = 8t, s_2 = 10t

10t = 8t + 80

2t = 80

t = 80 / 2 = 40 s

7 0
3 years ago
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