The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure.
The question is incomplete. Here is the complete question.
A floating ice block is pushed through a displacement vector d = (15m)i - (12m)j along a straight embankment by rushing water, which exerts a force vector F = (210N)i - (150N)j on the block. How much work does the force do on the block during displacement?
Answer: W = 4950J
Explanation: <u>Work</u> (W), in physics, is done when a force acts on an object that has a displacement form a place to another:
W = F · d
As the formula shows, Work is a scalar product, i.e, it results in a number, so, Work only has magnitude.
Force and displacement for the ice block are in 2 dimensions, then work will be:
W = (210)i - (150)j · (15)i - (12)j
W = (210*15) + (150*12)
W = 3150 + 1800
W = 4950J
During the displacement, the ice block has a work of 4950J
Answer:
The net force acting on the tennis ball while it is in contact with the racquet is 50.73 N
Explanation:
The impulse-momentum theorem said that the net impulse is equal to the change of the momentum, this is:
but the net impulse is too the net force times the change in time:
so using (2) on (1) we have:
Decomposing that on x and y components:
(See figure below) with Pfx = m*vfx= m*vf*cos(15°)=(0.058kg)(40m/s)cos(15°),
Pox = -m*vox= m*vo*cos(15°)=-(0.058kg)(30m/s)cos(15°), the same analysis to Pfy and Poy gives
Pfy=(0.058kg)(40m/s)sin(15°), Poy=-(0.058kg)(30m/s)sin(15°), using those values on (3) and (4) and solving for Fy and Fx:
So the net force acting on the tennis ball while it is in contact with the racquet is:
The angular speed of the minute hand of a watch is:
-- one revolution per hour
-- 2 pi radians per hour
-- 360 degrees per hour
-- 6 degrees per minute
-- 0.1 degree per second
Answer:
Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = qV, where q is the charge moved and V is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is
P
=
P
E
t
=
q
V
t
.
Recognizing that current is I = q/t (note that Δt = t here), the expression for power becomes
P = IV
Electric power (P) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅V= 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power P = IV = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (1 kA ⋅V = 1 kW). To see the relationship of power to resistance, we combine Ohm’s law with P = IV. Substituting I = V/R gives P = (V/R)V=V2/R. Similarly, substituting V = IR gives P = I(IR) = I2R. Three expressions for electric power are listed together here for convenience:
P
=
IV
P
=
V
2
R
P
=
I
2
R
.
Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, P can be the power dissipated by a single device and not the total power in the circuit.) Different insights can be gained from the three different expressions for electric power. For example, P = V2/R implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in P = V2/R, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.