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faltersainse [42]

3 years ago

10

Read the following excerpt about water availability to living organisms.

2 answers:

Musya8 [376]3 years ago

6 0

Answer:

1 percent (A)

lapo4ka [179]3 years ago

5 0

Answer:

1 percent

Explanation:

It says that only 3 percent of the water is fresh. So it can be 1 percent or 3 percent. But then it says that most of the water is locked up in glaciers and polar ice caps. So the animals would have a hard time getting to this water. So the rest is available for them. Approximately 1 percent is most reasonable.

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Ideally, rewards should be given immediately and frequently but

snow_lady [41]

The correct answer for the question that is being presented above is this one: "a. only from an instructor or supervisor." Ideally, rewards should be given immediately and frequently but <span>only from an instructor or supervisor to show authority. </span>

3 0

2 years ago

Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

Anon25 [30]

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>

This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

F is the applied force
v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

7 0

2 years ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

$A_{area}=0.140m^{2}\\ p_{air}=1.21 kg/m^{3}\\ m_{mass}=80kg$

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

$mg=1/2*C*p_{air}*v^{2}*A_{area}\\ v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s$

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

$v=(115*3.6)km/h\\v=414km/h$

5 0

3 years ago

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

The mass of the block is $m_b = 1.45 kg$

The spring constant is $k = 860 N/m$

The coefficient of static friction is $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

$F_y = m_b*g$

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

$F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

Now equating this two formulas and making x the subject

$x = \frac{m_b * g}{k * \mu}$

substituting values we have

$x = \frac{1.45 * 9.8}{860 *0.36}$

$x= 0.046m$

3 0

3 years ago

Someone please help !!

masha68 [24]

I’m pretty sure it’s c.... hope it helps and hope it’s right.

7 0

3 years ago