How do a neutron and a proton compare?

Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ign

iren [92.7K]

plane is flying at an altitude of 70 m

now if an object is dropped from it then time taken by object to drop on ground will be given as

$y = v_i* t + \frac{1}{2}at^2$

here initial speed in vertical direction must be zero as plane is moving horizontal

given that

y = 70 m

a = 9.8 m/s^2

$70 = 0 + \frac{1}{2}*9.8*t^2$

$t = 3.77 s$

now since the plane is moving horizontally with speed v = 44 m/s

so the horizontal distance moved by the object will be

$d = v_x * t$

$d = 44 * 3.77$

$d = 166.3 m$

so the distance moved by the box is 166.3 m

3 0

3 years ago

Read 2 more answers

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

Vikki [24]

Answer:

The force constant is $k =1.316 *10^{7} \ N/m$

The energy stored in the spring is $E = 1.68 *10^{7} \ J$

Explanation:

From the question we are told that

The mass of the object is $M = 4.8*10^{5} \ kg$

The period is $T = 1.2 \ s$

The period of the spring oscillation is mathematically represented as

$T =2 \pi \sqrt{ \frac{M}{k}}$

where k is the force constant

So making k the subject

$k = \frac{4 \pi ^2 M }{T^2}$

substituting values

$k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}$

$k =1.316 *10^{7} \ N/m$

The energy stored in the spring is mathematically represented as

$E = \frac{1}{2} k x^2$

Where x is the spring displacement which is given as

$x = 1.6 \ m$

substituting values

$E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2$

$E = 1.68 *10^{7} \ J$

7 0

3 years ago

A new prototype cup has been designed to keep liquids, such as hot coffee or cold annk, near their original temperature for long

san4es73 [151]

Answer:

hope this was good for u and I believe it would be solid

4 0

3 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

3 years ago

Convert time from 12-hour to 24-hour clock.

svet-max [94.6K]

Answer:

4:28

Explanation:

4:28am

a 12 hour clock continues going up after 12 (1:00pm=13:00). minutes stay the same. 12:00pm=00:00. this shows 4:28am, so you count 4 after 00:00.

6 0

3 years ago