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kkurt [141]
3 years ago
9

A 60 kg student is standing in the train station next to her 10 kg suitcase when her train is called. (A) Estimate how much work

she does picking up the suitcase. (B) She sprints 2 m, accelerating herself and her suitcase at 0.1 m/s2. How much force does she exert? (C) How much work does she do in this 2 m? (D) She now continues running at constant speed for 10 m to the train. How much work does she do in this 10 m? (E) What is her final kinetic energy as she reaches the train?
Physics
1 answer:
ASHA 777 [7]3 years ago
5 0

Answer

given,

mass of student = 60 Kg

mass of suitcase = 10 Kg

a) Work done by picking of Suitcase is equal to zero

b) acceleration = 0.1 m/s²

   distance = 2 m

Using second law conservation

F = m a

F = 10 x 0.1 = 1 N

c) Work done

  W = F x s

  W = 1 x 2 = 2 J

d) When the are moving with constant speed acceleration is equal to zero

F = m a

F = 10 x 0 = 0 N

W = F x s = 0 x s = 0 J

e)   work done = change in kinetic energy

        K.E = 2 J

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Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
Charra [1.4K]

Answer:

Time of flight  A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight  A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight  A is greatest .

8 0
3 years ago
7. If 8 million kg of water flows over Niagara Falls each second, calculate the power available at the bottom of the falls.
Alexxx [7]

Answer:

The power will be "3.92×10⁹ Watts". A further explanation is given below.

Explanation:

The given values as per the question,

Rate,

= 8 million kg

Distance,

= 50 m

Gravity,

= 9.8 m/s²

As we know,

The power will be:

⇒ Power = Rate\times Distance\times  Gravity

On putting the values, we get

⇒             =  8\times 10^6\times 50\times 9.8

⇒             =3.92\times 10^9 \  Watts

7 0
2 years ago
Please help this is physics!!!
larisa [96]
I’m pretty sure u have it right
8 0
2 years ago
How many joules of work are done against a truck when a force of 50 N pushes it 1 kilometer away
ELEN [110]

Answer:

Work = 50,000 J

Explanation:

Work = force * distance

Given that,

  • force = 50N
  • Distance = 1km = 1000m

Work = 50 * 1000

Work = 50,000 J

8 0
2 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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