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slavikrds [6]
3 years ago
12

Elements that typically give up electrons

Physics
1 answer:
Naddika [18.5K]3 years ago
8 0
Same answer as the first one above
You might be interested in
Half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value after
makkiz [27]

Answer:

44 years

Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44

3 0
3 years ago
At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi
Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0
3 years ago
A simple pendulum is made from a 0.54-m-long string and a small ball attached to its free end. The ball is pulled to one side th
Serga [27]

Answer:

0.37sec

Explanation:

Period of oscillation of a simple pendulum of length L is:

T = 2 π × √ (L /g)

L=length of string 0.54m

g=acceleration due to gravity

T-period

T = 2 x 3.14 x √[0.54/9.8]

T = 1.47sec

An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.

The ball will first have V(max) at T/4,

=>V(max) = 1.47/4 = 0.37 sec

3 0
3 years ago
I need help in my physics class and show me how it’s done
Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)

Where | A | is the magnitude of the vector and \alpha is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle \alpha.

Vector A is in the 4th quadrant

So:

A_x = 6\\\\A_y = -6.5

So:

|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846

So:

Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)

\alpha = -47.28 ° +360° = 313 °

\alpha = 313 °

Option 4.

4 0
3 years ago
Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa
hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}

Put the value into the formula

T=1\sqrt{\dfrac{1+0.6}{1-0.6}}

T=2\ years

Similarly,

The expression for distance cover by A

D=d\sqrt{1-\dfrac{v^2}{c^2}}

D=12\sqrt{1-(0.6)^2}

D=9.6\ ly

We need to calculate the time

Using formula of time

t=\dfrac{D}{v}

t=\dfrac{9.6}{0.6}

t=16\ years

We need to calculate the signals received by twin A

Using formula for number of signals

n=\dfrac{t}{T}

Put the value into the formula

n=\dfrac{16}{2}

n=8\ signals

Hence, 8 signals received by twin A during the trip.

7 0
3 years ago
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