The diagram shows two different types of fossils from the

A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height o

Thepotemich [5.8K]

Answer:

The value is $KE_b =0.710 \ J$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 5.00 \ g = 0.005 \ kg$

The mass of the wood is $m_w = 900 \ g = 0.90\ kg$

The height attained by the combined mass is $h = 8.0 \ cm = 0.08 \ m$

Generally according to the law of energy conservation

$KE _b = PE_c$

Here $KE_b$ is the kinetic energy of the bullet before collision.

and $PE_c$ is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as

$PE_m = [m_b + m_w] * g * h$

So

$KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08$

=> $KE_b =0.710 \ J$

3 0

3 years ago

Imagine tying a string to a ball and twirling it around you. How is this similar to the moon orbiting Earth? In this example, wh

leonid [27]

The string kind of acts like gravity

5 0

3 years ago

What is equilibrium????

IrinaK [193]

Answer:

Explanation:

An equilibrium is a state in which opposing forces or influences are banned.

An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.

7 0

2 years ago

Read 2 more answers

A wave has the wavelength equal to 396 nm. what must be the frequency of this wave

Anvisha [2.4K]

This assumes that the wave has velocity c (is light).

7 0

2 years ago

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

2 years ago