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Andru [333]
3 years ago
12

A 10 g dime and a 200 g gold coin are dropped from the Empire State Building in

Physics
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

Explanation:

The both hit the ground at the same time. There is no air resistance and that includes the factors brought up in B.

C and D are both eliminated. If you dropped an 800 kg honda Fit it would hit the ground at the same time as the two coins. When you look at the formulas used to derive time, they look like

d = vi * t + 1/2 a* t^2

vi = 0

the distance to fell is the same for all three masses.

a is the same for all 3 masses.

So when you solve for t, there is no mention of the mass of the object falling. Therefore t is dependent only on the height.

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A member of the marching band tunes their trombone inside their school. When they walk out to the field the instrument goes out
Alona [7]

The best explanation is the <em>difference</em> between the inside <em>temperature</em> and the outside temperature.

If the player doesn't change his emboucher (muscles and position of his lips), then the pitch produced by the instrument depends only on the physical dimensions of its plumbing, and the speed of sound in the tube.

BOTH of those things change slightly when the temperature changes.

4 0
3 years ago
Read 2 more answers
Help I'll mark brainliest
lubasha [3.4K]

Answer:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

Explanation:

4 0
3 years ago
crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied​
Klio2033 [76]

Answer:

 F = 5226.6 N

Explanation:

To solve a lever, the rotational equilibrium relation must be used.

We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise

         F d₁ = W d₂

where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.

In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is

d₂ = 200 cm = 2 m

therefore the distance to the applied force is

          d₁ = L -d₂

         d₁ = 5 -2

         d₁= 3m

we clear from the equation

          F = W d₂ / d₁

          W = m g

          F = m g d₂ / d₁

we calculate

      F = 800 9.8 2/3

      F = 5226.6 N

4 0
3 years ago
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

 W=3420.84*0.25[2401-65536]

 W=17084965.5J

 W=17085KJ

 

'

'

4 0
3 years ago
If you weighted 130 lbs on Earth how much would you weigh on the moon?
In-s [12.5K]

Answer:

21

Explanation:

Weight on the moon is 16.5 % of weight on earth

Weight on moon = 0.165 * 130

Weight on moon = 21 lbs

3 0
3 years ago
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