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3 years ago

What would you do with the force and mass to create a situation where the car has the highest acceleration possible?

Physics

1 answer:

olya-2409 [2.1K]3 years ago

3 0

Answer:

Increase the force and/or decrease the mass

Explanation:

A = F/m per Newton's second law, so increasing the force or decreasing the mass will increase acceleration

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A skydiver is preparing to jump out of a plane that is flying at a height of 2200 m. If the skydiver has a mass of 78 kg, what i

Illusion [34]

Answer:

1,681,680 J.

Explanation:

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3 years ago

NO LINKS! What are the definitions of initial horizontal velocity and initial vertical velocity? Please help! This like is my 10

dmitriy555 [2]

Answer:

Additionally, the initial horizontal velocity may be calculated by measuring the diameter d of the ball and dividing the result by the amount of time t required for the ball to pass across the photogate. Vo = d/t is the formula.

Initial vertical velocity is the vertical component of the initial velocity: v 0 y = sin 0 sin 0 = (30.0 m/s) sin 45° = 21.2 m/s

Explanation:

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8 0

3 years ago

Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to

Snowcat [4.5K]

Answer:

a) A1 = $\frac{\pi (d1)^{2} }{4}$

b) A1 = 2.688 cm $^{2}$

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = $\frac{A1 X v1}{v2}$

f) A2 = 0.7963cm $^{2}$

Explanation:

a) Area = $\pi r^{2}$

r = $\frac{d}{2}$

thus,

area = $\pi (\frac{d}{2})^{2}$

A1 = $\frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 = [tex]\frac{\pi (d1)^{2} }{4}$

A1 = $\frac{\pi (1.85)^{2} }{4}$

A1 = 2.688 cm $^{2}$

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = $\frac{Q1}{A1}$ = $\frac{860}{2.688}$ = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = $\frac{A1 X v1}{v2}$

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = $\frac{2.688 X 3.1994}{10.8}$ = 0.7963cm $^{2}$

3 0

4 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$