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2 years ago

What two characteristics are used to classify air masses

Physics

1 answer:

IgorLugansk [536]2 years ago

7 0

Answer: Temperature and humidity are the two characteristics used to classify air masses.

Explanation:

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If a car accelerates from rest at a constant 4 m/s

Readme [11.4K]

Answer:

The time it will take for the car to reach a velocity of 28 m/s is 7 seconds

Explanation:

The parameters of the car are;

The acceleration of the car, a = 4 m/s²

The final velocity of the car, v = 28 m/s

The initial velocity of the car, u = 0 m/s (The car starts from rest)

The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;

v = u + a·t

Where;

v = The final velocity of the car, v = 28 m/s

u = The initial velocity of the car = 0 m/s

a = The acceleration of the car = 4 m/s²

t = =The time it will take for the car to reach a velocity of 28 m/s

Therefore, we get;

t = (v - u)/a

t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s

The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.

4 0

2 years ago

A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th

vladimir1956 [14]

Answer:

Force, $F=2.27\times 10^4\ N$

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

$F.d=\dfrac{1}{2}m(v^2-u^2)$

Finally, it stops, v = 0

$F.d=-\dfrac{1}{2}m(u^2)$

$F=\dfrac{-mu^2}{2d}$

$F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}$

F = -22713.92 N

$F=2.27\times 10^4\ N$

So, the magnitude of the force that stops the bullet is $2.27\times 10^4\ N$

7 0

2 years ago

1) A plane lands on a runway with a speed of 125 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m

valentina_108 [34]

Answer: 1) a = 9.61m/s² pointing to west.

2) (a) Δv = - 37.9km/s

(b) a = - 6.10⁷km/years

Explanation: Aceleration is the change in velocity over change in time.

1) For the plane:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{125}{13}$

a = 9.61m/s²

The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.

2) Total change of velocity:

$\Delta v = v_{f}-v_{i}$

$\Delta v = -17.1-(+20.8)$

$\Delta v= -37.9$ km/s

The interval is in years, so transforming seconds in years:

v = $\frac{-37.9}{3.15.10^{-7}}$

$v=-12.03.10^{7}$ km/years

Calculating acceleration:

$a=\frac{-12.03.10^{-7}}{2.01}$

$a=-6.10^{7}$

Acceleration of an asteroid is a = -6.10⁷km/years .

5 0

2 years ago

A type of light bulb is labeled having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of

SashulF [63]

Answer:

$0.2592 \ or \ 25.92\%$

Explanation:

The exponential density function is given as

$f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t$

$\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}$

To find probability that bulb fails with the first 300hrs, we integrate from o to 300:

$P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}| \ 0\leq t\leq 300$

$P(0\leq X\leq 300)=-0.7408+1\\=0.2592$

Hence probability of bulb failing within 300hrs is 25.92% or 0.2592

3 0

2 years ago

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

3 years ago