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ser-zykov [4K]
2 years ago
7

Perhaps you have attended a party where balloons floated. A balloon filled with air does not rise above your head, but a balloon

filled with helium gas rises in the air to the end of its string. What does this tell you about the density of helium?
Physics
1 answer:
Drupady [299]2 years ago
3 0
Helium has a low dense than air
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3 years ago
An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 5 N, F2 = 8 N, and F3 = 5
DIA [1.3K]

The figure is missing: find it in attachment.

a) (3i - 5j) N

First of all, we have to write each force as a vector with a direction:

- Force F1 points downward, so along the negative y-direction, so we can write it as

F_1 = -5 j

- Force F2 points to the right, so along the positive x-direction, so we can write it as

F_2 = +8 i

- Force F3 points to the left, so along the negative x-direction, so we can write it as

F_3 = -5 i

Now we can write the net force, by adding the three vectors and separating the x-component from the y-component:

F=F_1+F_2+F_3 = -5j+8i-5i = (8-5)i+(-5)j=(3i-5j)N

b) 5.8 N

The magnitude of the net force can be calculated by applying Pythagorean's theorem on the components of the net force:

F=\sqrt{F_x^2+F_y^2}

where

F_x = 3 N is the x-component

F_y = -5 N is the y-component

Substituting into the equation,

F=\sqrt{(3)^2+(-5)^2}=5.8 N

c) -59.0^{\circ}

The angle of the net force, measured with respect to the positive x-axis (counterclockwise), can be calculated by using the formula

\theta=tan^{-1} (\frac{F_y}{F_x})

where

F_x = 3 N is the x-component

F_y = -5 N is the y-component

Substituting into the equation, we find

\theta = tan^{-1} (\frac{-5}{3})=-59.0^{\circ}

d) 0.84 m/s^2

The acceleration can be found by using Newton's second law:

F=ma

where

F is the net force on an object

m is its mass

a is the acceleration

For the object in the problem, we know

F = 5.8 N

m = 6.9 kg

Solving the equation for a, we find the magnitude of the acceleration:

a=\frac{F}{m}=\frac{5.8}{6.9}=0.84 m/s^2

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