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Andru [333]
3 years ago
13

17

Physics
1 answer:
ivann1987 [24]3 years ago
3 0

Answer:

Heat energy required (Q) = 70,537.5 J

Explanation:

Given:

Mass of water (m) = 225 grams

Initial temperature of water =  25.0°C

Final temperature of water = 100.0°C

Specific heat of water (c) = 4.18 J/g

Find:

Heat energy required (Q)

Computation:

Change in temperature Δt = Final temperature of water - Initial temperature of water

Change in temperature Δt = 100 - 25

Change in temperature Δt = 75°C

Using Heat energy formula

Q = (m)(c)(Δt)

Heat energy required (Q) = (225)(4.18)(75)

Heat energy required (Q) = 70,537.5 J

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The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo
spin [16.1K]

Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

W=\vec F\ \vec r

It can be written in its scalar version as

W=F.d.cos\theta

Being F and d the magnitudes of the force and displacement, and \theta the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

W=N.d.cos90^o=0

The work is zero because the force and the displacement are perpendicular

The static friction force exists only when the object is being applied a force of a magnitude not large enough to produce movement, i.e. the object is at rest. If the object is moved, the friction force is still present, but it's called dynamic friction force, usually smaller than the static.

Since in this case, there is no displacement, d=0, and the work is

W=F_r(0)cos180^o=0

3 0
4 years ago
A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t
Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

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3 years ago
"Younger teens often brag about a lack of sleep, but that usually stops when they get older. Why do you think that is?"
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3 years ago
A basketball player drops a 0.4-kg basketball vertically so that it is traveling at 5.7 m/s when it reaches the floor. The ball
Alik [6]

Answer:

(a) p = 3.4 kg-m/s (b) 37.78 N.

Explanation:

Mass of a basketball, m = 0.4 kg

Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)

It rebounds upward at a speed of 2.8 m/s  (as it rebounds so positive)

(a) Change in momentum = final momentum - initial momentum

p = m(v-u)

p = 0.4 (2.8-(-5.7))

p = 3.4 kg-m/s

(b) Impulse = change in momentum

Ft = 3.4

We have, t = 0.09 s

F=\dfrac{3.4}{0.09}\\\\F=37.78\ N

Hence, this is the required solution.

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Answer:

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Explanation:

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