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2 years ago

What is the equivalent resistance of a circuit that contains four 75.0 resistors connected in series to a 100.0 v battery

2 answers:

6 0

The equivalent resistance of several resistors connected in series is
the sum of their individual values.

Four 75-ohm resistors connected in series are electrically equivalent
to a single resistor of 300 ohms.

The battery voltage doesn't matter. In fact, it doesn't matter whether the
resistors are anywhere near a battery or not. They could just as well still
be in the little Radio Shack envelope on the top shelf of the storage room.
If they're connected in series, then they're still electrically equivalent to a
single 300-ohm resistor.

sergiy2304 [10]2 years ago

6 0

Answer: 300.0

Explanation:

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A train goes up a hill with a 15º incline. If the train has constant speed of 22 m/s, what are the vertical and horizontal compo

Kobotan [32]

Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.

8 0

2 years ago

Find the kinetic of a 0.1 kilogram toy truck moving at a speed of 1.1 meters per second

omeli [17]

KE = kinetic energy
PE = potential energy
GPE = gravitational potential energy
energy is always measured in Joules (J)

KE = (0.5) times the mass times the velocity^2
square the velocity first

Mass = (KE x 2) / v^2
square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47

GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height

height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
don't forget the unit of energy is in Joules

5 0

2 years ago

A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at

Jobisdone [24]

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = <em>360 km-m/s </em>

The direction of the change is <em>up /\ </em>.

8 0

2 years ago

An airplane weighing 5000 lb is flying at standard sea level with a velocity of 200 mi/h. At this velocity the L/D ratio is a ma

saul85 [17]

Answer:

98.15 lb

Explanation:

weight of plane (W) = 5,000 lb

velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s

wing area (A) = 200 ft^{2}

aspect ratio (AR) = 8.5

Oswald efficiency factor (E) = 0.93

density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}

Drag = 0.5 x ρ x $v^{2}$ x A x Cd

we need to get the drag coefficient (Cd) before we can solve for the drag

Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)

where

induced drag coefficient (Cdi) = $\frac{Cl^{2} }{n.E.AR}$ (take note that π is shown as n and ρ is shown as $p$ )

where lift coefficient (Cl)= $\frac{2W}{pAv^{2} }$ = $\frac{2x5000}{0.002377x200x293.3^{2} }$ = 0.245

therefore

induced drag coefficient (Cdi) = $\frac{Cl^{2} }{n.E.AR}$ = $\frac{0.245^{2} }{3.14x0.93x8.5}$ = 0.0024

since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)

Cd = 0.0024 + 0.0024 = 0.0048

Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.

Drag = 0.5 x ρ x $v^{2}$ x A x Cd

Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb

8 0

2 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

2 years ago

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