Lead-202 has a half-life of 53,000 years. How long will it take for 15/16 of a sample of lead-202 to decay?

A string is attached to a vibrating machine that has a frequency of 120 Hz. The other end of the string is passed over a pulley

BabaBlast [244]

Answer:

Explanation:

Given that,

The frequency of vibration is 120Hz

The mass of object attached is 0.5kg.

We want to find the tension In the string

The tension in the string is the weight of the object, it is how much gravity is pulling the object to the centre of the earth

Using newton second law.

F_net = m•a_y

The body is not accelerating the y-direction, then, a_y = 0

F_net = 0

Force acting on the string is the weight of the object and the tension in the string

T - W = 0

T = W

Where weight is mass × gravity

W = mg

Then,

T = W = mg

T = mg = 0.5 × 9.81

T = 4.905 N

The tension in the string is 4.905 N

4 0

3 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

In order of fastest to slowest, how do sound waves travel through different mediums?

Nikitich [7]

Gases, liquids, solids

7 0

3 years ago

A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t

Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 12\ m}{(1.2\ s)^2}$

a = 16.67 m/s²

Now put the value of a in equation (1) as :

$q=\dfrac{ma}{E}$

$q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}$

q = 0.0000249 C

or

$q=2.49\times 10^{-5}\ C$

Hence, this is the required solution.

5 0

3 years ago

Three diffrent examples of accelerated motion

LekaFEV [45]

Answer:

The three different examples of the accelerated motion are Falling/dropping of ball, Standing in circular rotating space, moving around the circle.

Explanation:

Acceleration is the change in velocity, which is related to the speed and direction in which the object is travelling. Hence, speeding up, slowing down and turning are few types . A simple example would be dropping a ball: as it falls its speed increases, which is a type of acceleration. A more complicated example would be standing in a circular, rotating space station. A point on the station moves in a circle, meaning that as it travels it must be turning (to remain in circular motion) making this another example of acceleration

3 0

3 years ago