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lisabon 2012 [21]

3 years ago

14

How much work does the electric field do in moving a proton from a point with a potential of +130 v to a point where it is -70 v

? express your answer in joules?

1 answer:

NikAS [45]3 years ago

6 0

Hi Gonnah,
W=qΔV
W= q(-70-130)
q of proton look it up and plug in and find out
it should be negative since you are doing work form greater potential to lower

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A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0

4 years ago

The note created by a flute will increase the speed of sound increases. When a marching band goes outside on a cold day, what wo

alexgriva [62]

A).

It would decrease because the speed of sound and temperature are proportional.

4 0

3 years ago

Read 2 more answers

A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe

tekilochka [14]

Answer:

Angular momentum, $L=6.47\times 10^{-3}\ m$

Explanation:

It is given that,

Radius of the axle, $r=3.21\ mm=3.21\times 10^{-3}\ m$

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

$\tau=\dfrac{dL}{dt}$

Torque acting on the top is given by :

$\tau=F\times r$

Here, F is the tension acting on it. Torque acting on the top is given by :

$\tau=2F\times r$

$2T\times r=\dfrac{L}{t}$

$L=2T\times r \times t$

$L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32$

$L=6.47\times 10^{-3}\ m$

So, the angular momentum acquired by the top is $6.47\times 10^{-3}\ m$ . Hence, this is the required solution.

7 0

4 years ago

If a one-pound weight were one foot from the pivot or shaft of a lever, how many pound-feet of force would result? A. 4 B. 1 C.

mixas84 [53]

The force is still 1 pound.

The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)

8 0

3 years ago

Read 2 more answers

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

6 0

3 years ago