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Katarina [22]
3 years ago
9

What is the equivalent resistance of the circuit?

Physics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

(B) 80 ohms

Explanation:

Since the circuit is a series connection, simply add the two resistances together to get 80 ohms.

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A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte
Zinaida [17]

Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

Thus,

total momentum of car before collision = 2000 x 3 = 6000

Also,

mass of sack = 500

mass of car and sack after collision = 500 + 2000 = 2500

velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

v = 2.4

the speed of the flatcar is 2.4 m/s

6 0
11 months ago
a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at th
Elza [17]
... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m

(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry

... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
4 0
3 years ago
Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di
S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F\alphae

F=ke

k=f/e.........*

where k is the force constant or the constant of proportionality

k=f/e

f_{eff} =f_{1} +f_{2}............................1

also for effective force constant

divide all through by extension

1) Total force is

Ft=F1+F2

Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

Substituting e1 and e2, you get

1/keq=1/k1+1/k2

Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.

4 0
3 years ago
1. A wave has a wavelength of 5 m and a frequency of 2 Hz. At what speed/velocity does the
yuradex [85]

Answer:

1.10m/s

2.0.1m

3.5Hz

Explanation:

v=velocity, f=frequency and T=wavelength

1.v=ft

v=2x5

=10m

2.v=ft

100=1000T

divide both sides by 1000

T=0.1m

3.v=fT

25=5f

divide both sides by 5

f=5Hz

4 0
2 years ago
What is the relationship between the angle of an incline and the acceleration of an object moving down the incline? How would yo
iren2701 [21]

Answer:

See Explanation

Explanation:

The relationship between angle of an incline and the acceleration of an object moving down the incline.

As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object

Parallel force = mgsin, perpendicular = mgcosΘ

With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.

mgsinΘ = ma

Fnet = ma

B.) From Fnet = ma

Fnet = ma

a = Fnet / m

Where Fnet = Net force = mgsinΘ, a = acceleration

5 0
3 years ago
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