Which term describes the time it takes for half of a radioactive elements atoms to decay?

A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.

elena-14-01-66 [18.8K]

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

α = angular acceleration

I = moment of inertia

I = MR² for a circular hoop

Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

6 0

3 years ago

Read 2 more answers

One kilogram-meter per second squared is also equal to what unit

yan [13]

Metres/second² is acceleration. maybe a bit more clarification?

8 0

2 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m

forsale [732]

Answer:

Acceleration due to gravity will be $g=5.718m/sec^2$

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing $=\frac{145}{100}=1.45sec$

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

$T=2\pi \sqrt{\frac{l}{g}}$

So $1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}$

$\sqrt{\frac{0.55}{g}}=0.230$

Squaring both side

${\frac{0.3025}{g}}=0.0529$

$g=5.718m/sec^2$

So acceleration due to gravity will be $g=5.718m/sec^2$

3 0

3 years ago

the illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the saurce?

Lelechka [254]

L = illuminance
A = surface
i = intensity

L = i / A ==: i = L * A

i = 6 lux * 4 m^2 = 24 lumen

8 0

2 years ago

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N

3 0

3 years ago