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3 years ago

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An electric heating element is connected to a 110v circuit and a current of 3.2 a is flowing through the element. How much energ

y is used up during a perAn electric heating element is connected to a 110v circuit and a current of 3.2 a is flowing through the element. [ How much energy is used up during a per - 1,760 Wh ]

2 answers:

Kisachek [45]3 years ago

7 0

2.580 Wh.... .........

Mashcka [7]3 years ago

3 0

110V x 3.2A = 352W

352W x 5h = 1,760Wh

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A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

lbvjy [14]

Answer:

$a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}$

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\$

b)

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}$

8 0

3 years ago

you ride a bike for 2 hours at 25 km/hr then 3 hours at 34 km/hr. what is your average velocity (in km/hr)?

mamaluj [8]

<h3>Answer:</h3>

30.4 km/hr

<h3>Explanation:</h3>

<u>We are given</u>;

Speed in the first 2 hours as 25 km/hr
Speed in the next 3 hours as 34 km/hr

We are required to determine the average velocity in km/hr

To get the average velocity we divide total distance by total time.
Thus, we need to determine the total distance

Distance = Speed × time

Distance covered in the first 2 hours;

= 25 km/hr × 2 hours

= 50 km

Distance in the next 3 hours

= 34 km/hr × 3 hours

= 102 km

Therefore, total distance = 50 km + 102 km

= 152 km

Total time = 2 hrs + 3 hrs

= 5 hours

Therefore;

Average speed = 152 km ÷ 5 hours

= 30.4 km/hr

Thus, the average speed is 30.4 km/hr

4 0

3 years ago

What item can be arranged to transform electrical energy to mechanical energy

meriva

Answer:

battery iron wire

Explanation:

7 0

3 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

3 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

<u>Now change in internal energy:</u>

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

B.

<u>Now heat added to the system:</u>

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

7 0

3 years ago