Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom

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

2 years ago

A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu

bija089 [108]

Answer:

J=36221 Kg.m/s

Explanation:

Impulse-Momentum Theorem

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

$p=mv$

The initial and final momentums are, respectively

$p_1=mv_1,\ p_2=mv_2$

The change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It is numerically equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

We are given

$m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s$

The impulse the car experiences during that time is

$J=2498(2.6-17.1)=2498(-14.5)$

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0

3 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

3 0

3 years ago

A healthy environment depends on biogeochemical cycles maintaining a balance of input and output. When forests are cut down, whi

Ann [662]

The correct answer will be c

8 0

2 years ago

Read 2 more answers

Sound waves in air travel at approximately 330m/s. Calculate the frequency of a 2.5m-long sound wave.

dusya [7]

Frequency = speed ÷ wavelength
= 330m/s ÷ 2.5m
= 132 Hz

8 0

3 years ago