All categories

1 year ago

What time is the eclipse happening tonight?.

1 answer:

3 0

The first location to see the partial solar eclipse begin is at 3.58 a.m. EST (08:58 UTC), the greatest point of total solar eclipse occurs at 6 a.m. EST (11:00 UTC) and the last location to see the partial eclipse end is at 8:02 a.m. EST (13:02 UTC) according to Time and Date.

You might be interested in

A student adds two vectors of magnitudes 48 m and 22 m. What are the maximum and minimum possible values for the resultant of th

Julli [10]

Answer:

<em>Maximum=70 m</em>

<em>Minimum=26 m</em>

Explanation:

<u>Vector Addition </u>

Since vectors have magnitude and direction, adding them takes into consideration not only the magnitudes but also their respective directions. Two vectors can be totally collaborative, i.e., point to the same direction, or be totally opposite. In the first case, the magnitude of the sum is at maximum. Otherwise, it's at a minimum.

Thus, the maximum magnitude of the sum is 48+22 = 70 m and the minimum magnitude of the sum is 48-22= 26 m

4 0

2 years ago

The rocket's acceleration has components $a_{x}(t)= \alpha t^{2}$ and $a_{y}(t)= \beta - \gamma t$, where \(\alpha = 2.50 {\

lbvjy [14]

it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]

5 0

3 years ago

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0

2 years ago

Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The

umka21 [38]

Find the below attachment

7 0

3 years ago

What forces are acting when you try to crack<br> an egg in your palm(will give brainliest)

Aleks [24]

Answer:

Kinetic energy in your palm

4 0

2 years ago