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Usimov [2.4K]
1 year ago
15

For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

ds the squirrel runs 2 m/s back to where he started. (A) How long does it take for the squirrel to get back to where he started?
Physics
1 answer:
Gala2k [10]1 year ago
7 0

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

T=\dfrac{30}{2}\ s= 15 \ s

Therefore, total time take to  get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

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A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to
erastovalidia [21]
Would presume you are asked to find the volume, since there is no second volume.

By General Gas Law:

P₁V₁/T₁ = P₂V₂/T₂

1.6 * 168 /255 = 1.3*V₂/285

V₂ = 1.6 * 168 * 285 / (1.3*255)

V₂ = 231.095

Final volume ≈ 231 cm³
7 0
2 years ago
Read 2 more answers
A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f
RSB [31]

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break  fly off = 120°

α = β = γ  = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}

A,B and C is momentum of the fragments

\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}

4 x V₂ = 2 x 3

V₂ = 1.5 m/s

3 0
1 year ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the
svet-max [94.6K]

Answer:

longer than

Explanation:

given,

time of nap = 10 min

speed of orbiting earth = 8000 m/s

c is the speed of light

using the equation of time dilation

t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}

now inserting all the values

t' = \dfrac{10}{\sqrt{1-\dfrac{8000^2}{3\times 10^8)^2}}}

t' = \dfrac{10}{0.9999}

t' = 10.001 s

on solving the above equation we will get a value greater than 10minutes.

hence, On earth time of nap measured will be longer than 10 min

3 0
2 years ago
A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of
qaws [65]

Answer:

\frac{E_1}{E_2}= 4

Explanation:

Capacitance C is given by

C= \frac{\epsilon_0A}{d}

A= area of capacitor cross section

d= distance

therefore,

C_1= \frac{\epsilon_0A_1}{d_1}

A_1= πR^2

d_1= d

C_2= \frac{\epsilon_0A_2}{d_2}

A_= π(2R)^2

d_2 = 2d

q= \frac{\epsilon_0A_1}{d_1}V_1

threfore

V_1= \frac{qd_1}{\epsilon_0A_1}

and

V_2= \frac{qd_2}{\epsilon_0A_2}

also we know that E= V/d

⇒\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}

⇒\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}

= A_1/A_2= \frac{4R^2}{R^2}=4

therefore,

\frac{E_1}{E_2}= 4

5 0
1 year ago
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