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Mashutka [201]
3 years ago
9

A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon

is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Physics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

The answer is 3.48 seconds

Explanation:

The kinematic equation

y= y0+V0*t+1/2*a*(t*t)

-50=0+(0)t+1/2(-9.8)*(t*t)

t=3.194 seconds

During ribbons ball,

x=x0+ Vt+1/2*a*(t*t)

x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)

x= 47.9157m

So, distance (D) = 100-47.9157= 52.084m

52.084m=0+15(t)+1/2*(0)(t*t)

t=52.084/15=3.472286= 3.48seconds

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You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is
Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

<u>Weight in air</u>

W (air) = mg, where g is the gravitational acceleration

<u>Weight with submerged with only one mass</u>

m(s)g + Fb = mg + m(b)g, <em>consider this to be equation 1</em>

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, <em>consider this to be equation 2</em>

<u>equation 1 – equation 2 would give us</u>

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

<u>Substituting V into the above equation we get</u>

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0
3 years ago
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w
Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

4 0
3 years ago
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pickupchik [31]

Answer:

Explanation:

All substances have characteristic physical and chemical properties. Physical properties are those that can be observed with the senses without changing the identity of the substance. Chemical properties describe how a substance can be changed into a new substance. Physical and chemical properties, such as color, density, boiling point, solubility, conductivity, and flammability, A. are always different between substances. B. depend on the amount of the substance. C. do not depend on the amount of the substance. D. have the same values for all substances.

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3 years ago
Read 2 more answers
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ELEN [110]

Answer:

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Explanation:

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6 0
2 years ago
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