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dusya [7]
3 years ago
9

HELP PLEASE!!!

Physics
1 answer:
Leya [2.2K]3 years ago
3 0

For the first one 320

second

1200W

Data

R = 12 Ω ∆V = 120V I =? P =?

Solution:

According to Ohm’s law,

∆V = I R

I = ∆V / R  

= 120 / 12  

= 10 A

Power P = I ∆V  

= 10 x 120  

= 1200 W

Third

∆V = 120 V P = 60 W I =? R =?

Use the formula, P = I ∆V

I = P / ∆V = 60 / 120 = 0.5 A

∆V = I R

R = ∆V / I = 120 / 0.5 = 240 Ω

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Does the uphill or downhill direction matter to the speed of the marble, or is the height the only contributing variable?
USPshnik [31]
B down hill is the fastest
3 0
3 years ago
A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
postnew [5]

Answer:

r=15.53 nm

F=9.57\times 10^{-13}N

Explanation:

Lets take electron is in between iron and uranium

Charge on electronq_1= -1.602\times 10^{-19}C

Charge on ironq_2= 2\times 1.602\times 10^{-19}C

Charge on uraniumq_3= 1.602\times 10^{-19}C

We know that force between two charge

F=K\dfrac{q_1 q_2}{r^2}  

K=9\times 10^9\dfrac{N-m^2}{c^2}

For equilibrium force between electron and iron should be force between electron and  uranium

Lets take distance between electron and  uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}  

K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}  

q_2= 2\timesq_1,q_3=q_1

\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}

So r=15.53 nm

So force

F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}  

F=9.57\times 10^{-13}N

7 0
3 years ago
How do I do these? My teacher didn’t show us how.
melisa1 [442]

Explanation:

Displacement is simply the change in position.  So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m.  So the displacement is:

Δx =  -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m.  The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m.  The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s.  It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s.  So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet.  Unfortunately, this isn't as easy as looking for where the lines cross on the graph.  We need to calculate their displacements.  We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2.  Half of the area is below the x-axis, and half is above.  Without doing calculations, we can say the total displacement for this car is 0.  This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when.  For Car 1, the area under the curve is a triangle.  So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time.  Since the line has a slope of 1 and y intercept of 0, we know v(t) = t.  So:

Δx = ½ t²

Now look at Car 3.  The area under the curve is a rectangle.  So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for.  Where are the cars at this time?  Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

7 0
3 years ago
How much heat is gained by 1.0 gram of iron when its heated 15 degrees celsius
Firdavs [7]
Q = mass water x specific heat water x delta T. 
<span>714,000 = mass water x specific heat water x 30.
Substitute specific heat water and solve for mass water.</span>
3 0
3 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
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