What type of data do scientists use to predict tornadoes?

The angular quantities: angular displacement, angular velocity, angular acceleration, are defined in a very similar way to the l

Lena [83]

Answer: [tex]12.415 rad.s^{-1}[/tex]

Explanation: Angular velocity is the rate of change in angular displacement.

We know that:

Angular velocity, $\omega= \frac{\Delta \theta}{t}$ ....................(1)

where:

t= time

$\Delta \theta$ = angular displacement in radians

Given that:

t = 4.10 s

Δθ = 50.9 radian

Putting the respective values in eq. (1)

$\omega = \frac{50.9}{4.10}$

$\omega = 12.415 rad.s^{-1}$

4 0

3 years ago

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$

$v=\sqrt{2gl}$

Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

$u=3.13\ m/s$

The initial speed of the ball $u_{1}=3.13\ m/s$

The initial speed of the block $u_{2}=0$

(a). We need to calculate the speed of the ball after collision

Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0

4 years ago

3 help with physics!! ---------------- will give brainliest

Zinaida [17]

Answer:

Answer B

Explanation:

An increase in resistance makes it harder for the electric current to pass through

6 0

2 years ago

What is one way in which radioactive decay is used to benefit society?

snow_lady [41]

They are used to measure and map effluent and pollution discharges from factories and sewerage plants, and the movement of sand around harbours, rivers and bays. Radioactive materials used for such purposes have short half-lives and decay to background levels within days.

7 0

3 years ago

Read 2 more answers

in which kind of radioactive decay would the number of protons in the resulting nucleus be more than in the initial nucleus?

9966 [12]

Answer:

A related type of beta decay

Explanation:

4 0

2 years ago