HELP! Ammeters are placed on each branch of a parallel circuit. How will their readings compare?

In odd nuclei, what determines the final spin of the nucleus?

Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

$1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }$

Here,

$j=\frac{1}{2}$

and, L=1.

Fort parity,

$(-1)^{L}$

Put the value of L.

Parity will be -1.

Now, spin will be

$S=(\frac{1}{2} )^{-1}$ .

7 0

3 years ago

As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce

WINSTONCH [101]

Answer:

a) x₁ = 290.50 feet , x₂ = 169.74 feet , b) v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis fr = m a

Y Axis N-W = 0

N = W = mg

The force of friction has the expression

fr = μ N

We replace

μ mg = ma

a = μ g

g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ a (feet / s2)

0.599 19.168

0.536 17,152

0.480 15.360

0.350 11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

v² = v₀² - 2 a x

When the speed stops it is zero

x₁ = v₀² / 2 a₁

Let's reduce speed

v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

x₁ = 80.667² / (2 11.2)

x₁ = 290.50 feet

The minimum braking distance

x₂ = 80.667² / (2 19.168)

x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

0 = v₀² - 2 a x

v₀ = √2 a x

We calculate the speed for the two accelerations

v₀₁ = √ (2 11.2 155)

v₀₁ = 58.92 feet / s

v₀₂ = √ (2 19.168 155)

v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0

3 years ago

A power plant supplies 1,100 megawatts of power to the electric grid. how many joules of energy does it supply each second?

slavikrds [6]

The answer is it will supply 1.1 x 10⁹ J of energy each second.
we can calculate this by using the following equation;
P = W/t
W = P x t
and by work energy relation;
E = W = P x t
1 watt = 1j/s
1megawatt = 1000000 = 10⁶ j/s
E = 1100 x 106 x 1 
E = 1.1 x 10⁹ J

5 0

3 years ago

A small economy car (low mass) and a limousine (high mass) are pushed from rest across a parking lot, equal distances with equal

Studentka2010 [4]

Answer:

The car that receives more kinetic energy is the small economy car.

Explanation:

K.E = 0.5*mv²

Where;

K.E is the kinetic Energy

M is the mass of an object

V is the velocity of the moving object

But F = m(v/t), from Newton's second law of motion

If equal forces were applied to the two cars, then the velocity of each car will be calculated as follows.

v = (Ft/m)

v² = (Ft/m)²

Substitute in the value of v² into Kinetic energy equation

K.E = 0.5*mv²

K.E = 0.5*m(Ft/m)² = (0.5*F²t²)/m

Also assuming equal distance, equal force and assuming equal time for both cars.

The above equation will reduce to, K.E = k/m

Where k = 0.5*F²t², which is equal in both cars.

Thus, Kinetic energy will depend only on the mass of each car.

From the above expression, Kinetic Energy received by each car is inversely proportional to the mass of the car.

A small economy car (low mass) will receive more kinetic energy while a limousine (high mass) car will receive less kinetic energy.

Therefore, the car that receives more kinetic energy is the small economy car.

6 0

3 years ago

Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910 m in Yosemite

german

Answer:

The jumper is in freefall for 12.447 seconds.

Explanation:

Let's start by calculating how far the jumper falls.

Initial height (on cliff) = 910 m

Final height after freefall = 150 m

Distance the jumper falls in freefall = 910 - 150 = 760 m

We can now use the equation of motion below to solve for the time:

$s=u*t+\frac{1}{2} (a*t^2)$

here. acceleration = 9.81 m/s (due to gravity)

initial speed (u) = 0 m/s (because vertical speed is 0 at the start)

and distance (s) = 760 meters (as calculated above)

So for speed we get:

$760=0*t+0.5(9.81*t^2)$

$760=4.905t^2$

t = 12.447 seconds

5 0

3 years ago