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arlik [135]
3 years ago
15

A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?​

Physics
1 answer:
Trava [24]3 years ago
8 0

Answer:

23.49m

Explanation:

Distance =  velocity x time

8.7 x 2.7 = 23.49m

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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
3 years ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
3 years ago
A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?
Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

5 0
2 years ago
Read 2 more answers
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