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3 years ago

Write the derivation and unit of impulse

Physics

1 answer:

ivann1987 [24]3 years ago

7 0

Answer:

impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction. The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kg⋅m/s).

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A belief, judgment, or way of thinking about something is called a(n) __________.

NISA [10]

That is an opinion. Opinion is not fact, although it can be based upon fact.

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3 years ago

Read 2 more answers

A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. the dynamic coefficient of friction between the

Talja [164]

The first thing you should know is that the work is defined as:
W = F * d
Where
F = force
d = displacement
We have then
(a) the block
F = (0.2) * (100) = 20
d = 100
W = (20) * (100) = 2000 ft.lbf
(b) the man as the system.
F = (0.2) * (100 + 180) = 56
d = 100
W = (56) * (100) = 5600 ft.lbf
answer:
(a) 2000 ft.lbf
(b) 5600 ft.lbf

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3 years ago

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe

Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet = 327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

= 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

= 0.00353 × 128 × (327-40)

= 0.00353 × 128 × 287

= 129.68 J

Quantity of heat required to melt the bullet:

$Q = mH_f$

Q = 0.00353 × 24500

= 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

= 216.17 J

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3 years ago

Define , gravitational acceleration

Vitek1552 [10]

The simplest answer would be "acceleration due to gravity."

The exact value of this acceleration changes depending on which planet your on (for example).

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3 years ago

Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related

inessss [21]

<h2>Answer: 136.363 m</h2>

Explanation:

We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:

$c=\lambda.\nu$ (1)