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Ad libitum [116K]

3 years ago

6

A current of 1.75 A flows through a 55.4 Ω resistor for 9.5 min. How much heat was generated by the resistor? Answer in units of

J.

1 answer:

kirill115 [55]3 years ago

6 0

Answer:

$9.67\times 10^4 J$

Explanation:

We are given that

Current=I=1.75 A

Resistance =R=55.4 ohm

Time=t=9.5 min= $9.5\times 60=570 s$

1 min=60 s

We have to find the heat generated by the resistor.

We know that

Heat=E= $I^2Rt$

Using the formula

Heat, $E=(1.75)^2\times 55.4\times 570=9.67\times 10^{4} J$

Hence, the heat generated by the resistor = $9.67\times 10^4 J$

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Where is parallel circuits used?

suter [353]

Answer:

if one bulb burns out, the other bulbs in the fixture continue to operate. Other uses include an electronic OR gate, where two switches are in a parallel circuit: one of the switches must be closed for the circuit to function.

8 0

4 years ago

The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed

padilas [110]

Answer:

<em>Details in the explanation</em>

Explanation:

<u>Vertical Launch</u>

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

$h = -16t^2 + 64t$

<em>Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.</em>

We now have to evaluate h for the following times, assuming h comes in feet

At t=1 sec

$h = -16(1)^2 + 64(1)=64-16=48\ ft$

The object is at a height of 48 feet

At t=2 sec

$h = -16(2)^2 + 64(2)=128-64=64\ ft$

The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below

At t=3 sec

$h = -16(3)^2 + 64(3)=192-144=48\ ft$

The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down

At t=4 sec

$h = -16(4)^2 + 64(4)=256-256=0\ ft$

The object is at ground level and has returned to the launch point.

5 0

3 years ago

5. Annie drags her little red wagon with a mass of 5.00 kg, up a hill that has an angle of

maxonik [38]

Answer: $25.08\ m/s^2$

Explanation:

Given

mass of wagon m=5 kg

elevation

$\theta =30.9^{\circ}\\F=150\ N$

Sin component of weight will oppose the applied force therefore we can write

$F-W\sin \theta=ma\\where\\W=weight(mg)\\a=acceleration$

$150-5\times 9.8\times \sin (30.09)^{\circ}=5\times a\\125.43=5a\\a=25.08\ m/s^2$

6 0

3 years ago

Forces start and change the motion of an object. Question 1 options: True False

Vlad1618 [11]

I think it’s true I may be wrong tho

8 0

3 years ago

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago