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3 years ago

What can i find when im only given the initial and final speed of an object

Physics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

you can you should find the velocity of the thing by which it is running

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How do I do number 8, plz respond quick, I have a big unit test on this and I’m rocking a 65 science average rn. Thank you for t

xxTIMURxx [149]

Answer:

3 is the correct answer for number 8

6 0

3 years ago

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A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U

Lubov Fominskaja [6]

Answer:

Explanation:

W = Vq

q = CV

W = V.CV

W = CV²

W/C = V²

V = √(W/C)

√(W1/C1) = √(W2/C2)

√(u/c) = √(x/2c)

x = 2u

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3 years ago

A mother and daughter press their hands together and then push apart while ice skating. Immediately after they push away from ea

kherson [118]

Answer:

Explanation:

The daughter moves with greater acceleration backwards because of her weight.

7 0

3 years ago

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Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

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3 years ago