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Nadusha1986 [10]
2 years ago
10

Pliss help...

Physics
1 answer:
natita [175]2 years ago
7 0

Answer:

I think it is A

Explanation:

Below, what belongs to the organs of the circulatory system in humans is...

<em>a. heart - kidneys - lungs</em>

b. heart-blood vessels - blood

c. heart - kidney - liver

d. heart - lungs

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A vector has an x component of 24.0 units and the magnitude of the vector is 25 units. Find
GrogVix [38]

Answer:

Magnitude of y-component=7 units

Direction= angle of vector A with x-axis= 16.26^{o}

Explanation:

Explained solution is in the picture attached

4 0
2 years ago
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

7 0
3 years ago
Which part of a stream's sediment load moves the slowest?
Rudiy27
The bed load moves the slowest from all the parts of the stream's sediment. It consists of particles suspended that are suspended and float around the bed. This part is the slowest in motion, as it rolls, and moves with the flow. The particles near the bed are not dissolved so they settle at the bottom and move with the stream.
4 0
3 years ago
Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec
SOVA2 [1]

Answer:

F_{net} = 6.879\times 10^{- 7}\ N

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

F = \frac{GMm}{\frac({d}{2}^{2})}

F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

F = \frac{GMm}{\frac({d}{2}^{2})}

F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

F_{net} = F + F'

F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N

6 0
3 years ago
Pause
nirvana33 [79]

Answer:

C

Explanation:

a=f/m

10Kgm/s2/4kg

2.5m/s2

3 0
2 years ago
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