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3 years ago

The product PV, where P and V represent pressure and volume respectively, is a unit for which?

1 answer:

6 0

PV is actually energy. P = F/A force per area, and V = A L, so PV = F L and force times distance is work which is energy. If you have P in N/m^2 and V in m^3, you have Joules, N-m.

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Answer:

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Explanation:

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3 years ago

What is science and physics? yeha avo.

Dafna1 [17]

Answer:

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3 years ago

A stone on ground is zero energy

NNADVOKAT [17]

Answer:

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3 years ago

What is the x-component of the force on the charge located at x = 8 cm given that q = 1.1 μC in N?

jonny [76]

Answer:

Answer is = 12.3 x 10^6 N

Explanation:

We need to calculate force on the charge which is in Newtons:

Formula is = K q1q2/r^2

k is constant depeneds on the permitivity of the medium and its value is 9x10^9 Nm^2/C^2

q1 and q2 are charges and r is the distance between these two charges so just by putting values we get F= 9x10^9 x 1.1x10^-6/(8x10^-2)^2 (we get the above answer in newton)

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4 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

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4 years ago