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Bezzdna [24]
3 years ago
9

Please break down a Pedigree chart..30 points because I'm generous :p

Physics
1 answer:
grigory [225]3 years ago
6 0
A pedigree chart is a chart that shows the occurrence of a organisms of generation, human, animals.
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A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.
dedylja [7]

Answer:

F = 768 N                  

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, \dfrac{dm}{dt}=240\ kg/min = 4\ kg/s

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

F=\dfrac{dp}{dt}

F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}

Since, the speed is constant,

F=m\dfrac{dv}{dt}

F=v\dfrac{dm}{dt}

F=3.2\times 240

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0
4 years ago
A car has a force of 0.27 N and an acceleration of 3m/s2 What is the car's mass?​
lyudmila [28]

Answer:

0.09 kg

Explanation:

f=0.27

a=3

m=f/a

m=0.27/3

m=0.09

7 0
3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
I need help with 2 questions . Please help c:
dezoksy [38]
A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
Therefore the correct answer is "all of the above".

b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
The correct answer is again "all of the above".
8 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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