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3 years ago

What happens when two forces act in the same direction?

Physics

2 answers:

SSSSS [86.1K]3 years ago

7 0

The two forces are added when in hook ground the same direction

Wewaii [24]3 years ago

5 0

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

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Please help on this one

Murljashka [212]

Just find the area of the graft

4m/s x 5s
=20m

7 0

3 years ago

Read 2 more answers

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

Taking test, need physics help<br> ASAP please!

Neko [114]

Answer:

one free advice don't believe all this free advices give din brainly ok

5 0

3 years ago

Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle

tangare [24]

Answer:

$F=1.159$

Explanation:

From the question we are told that:

Mass of pulley $M=1kg$

Radius $r=12cm$

Mass of block A $M_a=2.1kg$

Mass of block B $m_b=4.1kg$

Spring constant $\mu= 358 J/m2$

Generally the equation for Torque is mathematically given by

Since $\sumF=ma$

At mass A

$T_2-f_3=2.1a$

At mass B

$4.8-T_1=4.1a$

At Pulley

$R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}$

$R(T_1-T_2)=0.55a$

Therefore the equation for total force F

At mass A+At mass B+At Pulley

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a$

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a$

$-f_3+4.1=6.75a$

$-f_3=6.75a+4.8$

Since From above equation

$M_{eff}=6.7kg$

Therefore

$T=2\pi \sqrt{{\frac{M_{eff}}{k}}$

$T=2\pi \sqrt{{\frac{6.75}{\mu}}$

$T=0.862s$

Generally the equation for frequency is mathematically given by

$F=\frac{1}{T} \\F=\frac{1}{0.862}$

$F=1.159$

3 0

3 years ago

Why are the lunar mountains smoothly rounded rather than having sharp, pointed peaks (as they were almost always depicted in sci

omeli [17]

Answer:

Explained

Explanation:

Lunar mountains are smoothly rounded rather than having sharp, pointed peaks because the lunar moons are formed because of the impact of asteroids and comets on its surface whereas the Earth mountains are formed because of the surface activity of volcanos, collision of land masses and activity of plate tectonics that is why they have sharp and pointed tops.

6 0

3 years ago