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Gnoma [55]
3 years ago
14

Famed stunt pilot, Cleonvia Thread, pulls out rapidly from a dive. He is traveling at 222 mi/h at the bottom of his trajectory,

and at that instant is traveling on a curve of radius 820 ft. r v What acceleration does he
Physics
1 answer:
Simora [160]3 years ago
8 0

Answer:

0.39 m/s²

Explanation:

From the question,

a = v²/r.................... Equation 1

Where v = velocity, r = radius.

Given: v = 222 mi/h = (222×0.44704) m/s = 9.83 m/s, r = 820 ft = (820×0.3048) m = 249.94 m.

Substitute thses values into equation 1

a = 9.83²/249.94

a = 96.63/249.94

a = 0.39 m/s²

Hence the  acceleration is 0.39 m/s²

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Insulin and glucagon help maintain blood glucose levels

6 0
3 years ago
An ice cream truck is going 25m/s to the East. It accelerates to 45m/s in the same direction over 5s. What is its acceleration?
Naya [18.7K]

Hello!

We can use the kinematic equation:
a = \frac{v_f - v_i}{t}

a = acceleration (m/s²)

vf = final velocity (45 m/s)
vi = initial velocity (25 m/s)

t = time (5 sec)

Plug in the givens:
a = \frac{45-25}{5} = \frac{20}{5} = \boxed{4 m/s^2}

6 0
2 years ago
Skater A travels with velocity of 3.2 m/s and has a momentum of 200 kg m/s
Ivenika [448]

Here's link^{} to the answer:

bit.^{}ly/3gVQKw3

5 0
2 years ago
A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1
stellarik [79]

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = \frac{4}{3}\pi r^3

M = Molar mass of helium = 4.0026\times 10^{-3}\ kg/mol

\rho = Density of surrounding air = 1.19\ kg/m^3

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg

Mass of helium is 6.4356 kg

Moles of helium

n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489

Ideal gas law

P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0
3 years ago
What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed
RoseWind [281]
<h3><u>Minimum uncertainty in the vertical component of the momentum of each photon:</u></h3>

According to Heisenberg's Uncertainty principle, both the “position and velocity of the particle” cannot be measured exactly at the same time. The momentum of the particle equals the product of its mass and velocity. And it can be inferred that the “product of the uncertainties” in the “momentum and the position” of a particle equals \frac{h}{4 \pi}.  

Immediately after the photon has passed through the slit, given particle has a momentum uncertainty of \Delta P_{x} and its position uncertainty is \Delta x, then the minimum uncertainty in its momentum will be

\Delta P_{x} . \Delta x \geq \frac{h}{4 \pi}

8 0
3 years ago
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