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motikmotik
2 years ago
13

Find the average velocity of the object from point B to C.

Physics
1 answer:
ExtremeBDS [4]2 years ago
6 0
The average velocity of the object in moving from point-B to point-C is

                 (the straight-line distance and direction from point-B to point-C)
divided by
                 (the time the object takes to make the trip) .
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How can I solve this?
inysia [295]
There is no equation here
3 0
3 years ago
What would a series circuit be used for?
igor_vitrenko [27]

Answer:

C

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

6 0
2 years ago
Can someone explain how exactly to calculate the net force? Do you subtract the smaller number from the larger number? Any help
aleksley [76]

Explanation:

Let's say right is positive and left is negative.

F₁ = -150 N

F₂ = 220 N

Fnet = F₁ + F₂

Fnet = -150 N + 220 N

Fnet = 70 N

The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.

And since Fnet isn't 0, the force is unbalanced and the motion is changing.

3 0
3 years ago
Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1
kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex]

and v(0)=\hat{i}

r(0)=\hat{j}

we know a=\frac{\mathrm{d} v}{\mathrm{d} t}

\int dv=\int adt

v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt

v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c

at t=0

v(0)=0-1\cdot \hat{j}+0+c

c=\hat{i}+\hat{j}

v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}

and \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)

\int dr=\int vdt

r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt

r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2

at t=0

r(0)=\hat{j}

r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}

       

4 0
3 years ago
Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d
Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

h_{fg} = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

T_{L} = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

T_{H} = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, q_{reject} = h_{fg} = 171.82 kJ/kg

We know COP of heat pump

COP = \frac{T_{H}}{T_{H}-T_{L}}

        = \frac{304.31}{304.31-254.23}

         = 6.076

Therefore, Work out put, W = \frac{q_{reject}}{COP}

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

8 0
3 years ago
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