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Sveta_85 [38]
3 years ago
11

A 2.0-kg cart collides with a 1.0-kg cart that is initially at rest on a low-friction track. After the collision, the 1.0-kg car

t moves to the right at 0.50 m/s and the 2.0-kg cart moves to the right at 0.25 m/s . If the positive direction is to the right, what was the initial velocity of the 2.0-kg cart?
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

Initial velocity of first cart will be 0.5 m /sec

Explanation:

We have given mass of the cart m_1=2kg

Its collides with other cart of mass m_2=1kg

Let the initial velocity of cart 1 is v_{1i}

As the car 2 is initially at rest so v_{2i} = 0 m/sec

After the collision velocity of first cart is 0.25 m/sec

So v_{1f} = 0.25 m/sec and velocity of second cart is 0.50 m /sec

So v_{2f} = 0.5 m/sec

From conservation of momentum we know that

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

2\times v_{1i}+1\times 0=2\times 0.25+1\times 0.5

v_{1i} = 0.5 m /sec

So initial velocity of first cart will be 0.5 m /sec

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A point charge of -2 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m. Find the x and y coord
Soloha48 [4]

Answer:

x coordinate = -1.66 m

y coordinate is = -0.825m

Explanation:

Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12

We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2

0.0356s^2 -0.019s-0.0897=0  

s=1.876m

The angle of the line between the two charges is arctan(.5/1) = 26.6o

x coordinate = -1.876*cos(26.6) = -1.66m

y coordinate is -1.876*sin(26.6) = -0.825m

3 0
3 years ago
A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. what is the average power delivered by the engine? (1 hp = 746 w)
AVprozaik [17]
The solution for this is:
Work done = force * distance = m*a*d and power = energy/time 
The vo=0 and vf = 25 m/s and t=7 sec. This gives... 
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N. 
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again). 
5 0
3 years ago
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If I were to transmit a radio wave in our three dimensional world could a fourth dimensional “being” be able to receive it?
Morgarella [4.7K]
Depends. Are you talking about a mathematical 4th dimension (in which there is infinite dimensions) or some sort of etheral dimension (in which there is no scientific evidence for)

If you mean the first then yes. But it depends how these beings exist. From our understanding we only can theorize shapes in 4-d and if we assume that there is only one universe these "beings" arleady exist and thus any message in 3-d would be sent to them like a shadow ("flat").
If they exist in a alternate "plane" then you would need some method to transverse this plan and if u did, then we would easily be able to communicate, but we would at first sound like a wild animal. They either would ignore us, not understand or perceive us, or they would attempt to send back a signal (essential they are ET's)

IF you mean the second then thats some mystic stuff and its pretty creepy (although a fun read for me :P)
 
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5 0
3 years ago
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

Change in entropy \Delta s = m(s_2 - s_1)

              =3( 3.36035 - 2.88) =  1.44 kJ/kg

8 0
4 years ago
A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A,
taurus [48]

Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

8 0
3 years ago
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