A plane took 3.55 hours to finish a journey. If the distance of

An advertising balloon shaped like a giant soda can is 15 feet tall and 7 feet wide. How many cubic feet of helium will be neede

Iteru [2.4K]

Answer:

576.975 cubic feet of helium will be needed to fully inflate the balloon

Explanation:

To determine how many cubic feet of helium will be needed to fully inflate the balloon, we will determine the volume of the balloon.

From the question,

The balloon is shaped like a giant soda can.

A giant soda can is cylindrical.

Hence, we can determine the volume of the balloon shaped like a giant soda can by using the formula for finding the volume of a cylinder.

The formula for finding the volume of a cylinder is

V = πr²h

Where V is the volume of the cylinder

π is a constant (Take π = 3.14)

r is the radius of the cylinder

and h is the height of the cylinder

From the question, the balloon is 15 feet tall and 7 feet wide

Hence,

Height, h = 15 feet

Width = 7 feet

(NOTE: The width of a cylinder is the same as the diameter)

Then, diameter = 7 feet

Radius, r is given by

Radius = Diameter / 2

Then, Radius = 7 feet / 2 = 3.5 feet

∴ Radius, r = 3.5 feet

Now, for the volume of the balloon,

V = πr²h

V = 3.14 × (3.5)² × 15

V = 3.14 × 12.25 × 15

V = 576.975 cubic feet

This is the volume of the balloon.

Hence, 576.975 cubic feet of helium will be needed to fully inflate the balloon.

5 0

3 years ago

A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

Stells [14]

Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)= $1.31 m/s^2$

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

$5.47=0+1.31\times t$

$t=\frac{5.47}{1.31}=4.17 s$

Distance traveled during this time is

$s=ut+\frac{1}{2}at^2$

$s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m$

So remaining distance left to travel with constant velocity=153.57 m

thus time $=\frac{distance}{speed}$

$t_2=\frac{153.57}{5.47}=28.07 s$

Total time=28.07+4.17=32.24 s

6 0

3 years ago

A baseball player hits a baseball with a bat. The mass of the ball is 0.25 kg. The ball accelerates at 200 m/s2

EastWind [94]

Answer:

50N

Explanation:

Force (N) = mass (kg) × acceleration (m/s²)

0.25kg times 200m/s² = 50N

7 0

3 years ago

An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

So

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

So

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

8 0

3 years ago

I need help to figure out how to solve this problem and solve it!!!

wariber [46]

well it looks like the walk at a constant increasing pace then at a constant pace then increaseing pace then constant pace then they slow down then walk at a constant pace then walk at a constantly increasing pace

plz rate me brainliest

4 0

4 years ago

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