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Annette [7]
3 years ago
5

A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of

the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)
Physics
1 answer:
kirill [66]3 years ago
7 0

Answer:A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)

Explanation:

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You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out
kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0
3 years ago
Greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
maw [93]

{\large{\bold{\rm{\underline{Given \; that}}}}}

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

{\large{\bold{\rm{\underline{To\; find}}}}}

★ The speed of the hound and the hare

{\large{\bold{\rm{\underline{Solution}}}}}

★ The speed of the hound and the hare = 25:18

{\large{\bold{\rm{\underline{Full \; Solution}}}}}

\dashrightarrow  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

 But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

 Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

  • (5/3)a × 5

  • 25/3a metres

 Now the distance travelled by the hare in it's 6 leaps..!

  • 6a metres

 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

  • 25/3a = 6a

  • 25/3 = 6

  • 25:18
5 0
3 years ago
A boy weighs 400N. what is his mass?
Basile [38]

≈40.8kg


this is to take up space


m

=

W

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9.8

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9.8

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9.8

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m

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9.8

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9.8

m/s

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kg

m

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9.8

m/s

2


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m

=

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9.8

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40.8

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4 0
3 years ago
Read 2 more answers
Why does it seem that the Sun is rising and setting in the sky?
Anna [14]
B when is spins it means its rotating east and west directions
6 0
3 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

where

\phi = Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

\lambda = wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]

Magnetic field Equation

B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

Where B₀= E₀/c

B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T

B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]

6 0
3 years ago
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