A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.

The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed

padilas [110]

Answer:

<em>Details in the explanation</em>

Explanation:

<u>Vertical Launch</u>

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

$h = -16t^2 + 64t$

<em>Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.</em>

We now have to evaluate h for the following times, assuming h comes in feet

At t=1 sec

$h = -16(1)^2 + 64(1)=64-16=48\ ft$

The object is at a height of 48 feet

At t=2 sec

$h = -16(2)^2 + 64(2)=128-64=64\ ft$

The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below

At t=3 sec

$h = -16(3)^2 + 64(3)=192-144=48\ ft$

The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down

At t=4 sec

$h = -16(4)^2 + 64(4)=256-256=0\ ft$

The object is at ground level and has returned to the launch point.

5 0

3 years ago

If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant for

densk [106]

Answer:

Part A

The volume of water displaced is 4.1887902 × 10⁻³ m³

Part B

The buoyant force is approximately 40.93 N

Explanation:

From the question, we have;

The radius of the ball suspended (barely floating) in the water, r = 0.1 m

The density of the water, ρ = 997 kg/m³

Part A

The volume of the ball = The volume of a sphere = (4/3)·π·r³

∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³

Therefore;

The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³

The volume of water displaced, V = 4.1887902 × 10⁻³ m³

Part B

The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g

The buoyant force = m × g

Where;

g ≈ 9.8 m/s²

The mass of the water, m = ρ × V

∴ m = 997 kg/m³ × 4.1887902 × 10⁻³ m³ = 4.17622383 kg

The buoyant force = 4.17622383 kg × 9.8 m/s² ≈ 40.93 N.

7 0

3 years ago

A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns

Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

$f'=\frac{V+V_{o}}{V-V_{s}} f$ (1)

Where:

$f=250,000 Hz$ is the actual frequency of the sound wave

$f'=240,000 Hz$ is the "observed" frequency

$V=340 m/s$ is the velocity of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which is the balloon

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$ (2)

$V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}$ (3)

$V_{s}=-14.16 m/s$ (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ( $t=4.8 s$ ) given as data, we can find the distance:

$d=V_{s}t$ (5)

$d=(14.16 m/s)(4.8 s)$ (6)

Finally:

$d=68 m$ (8) This is the distance of the balloon from the radar station

6 0

3 years ago

Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat

fomenos

The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².

<h3 /><h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

P₂=ρ₂gh₂

P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625 N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ4

6 0

2 years ago

According to this graph, which statement describes how the amount of iron changes over time in the reaction

VashaNatasha [74]

The graph shows the production of Fe, from the graph that it increases rapidly and then slowly increases.<span>The answer is a! (:
</span>

5 0

4 years ago