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3 years ago

14

Describe two physical environmental factors in your environment at the moment

1 answer:

Anna11 [10]3 years ago

7 0

Answer:

The factors in the physical environment that are important to health include harmful substances, such as air pollution or proximity to toxic sites the focus of classic environmental epidemiology access to various health-related resources e.g. healthy or unhealthy foods, recreational resources, medical care.

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Difference between elastic and inelastic collisions

Eva8 [605]

An elastic collision<span> is where there is no loss of kinetic energy in the </span>collision<span>. Momentum is conserved saved in </span>inelastic collisions<span>, but cannot track the kinetic energy through the </span>collision<span> since some of it is changed into other forms of energy.</span>

3 0

3 years ago

What is the work of the force F (6.0N)(4.0N)j(-2.0N)k when the object moves from an initial point with coordinates (1.5 m, 3.0m,

Alina [70]

Answer:

option (c) - 10 j

Explanation:

F = (6 i + 4 j - 2 k) N

r1 = (1.5, 3, -4.5) m = (1.5 i + 3j - 4.5 k) m

r2 = (4, -2.5, - 3) m = (4 i - 2.5 j - 3 k) m

displacement, r = r2 - r1 = ( 2.5 i - 5.5 j + 1.5 k) m

Work done is defined as the dot product of force vector and teh displacement vector.

$W = \overrightarrow{F}.\overrightarrow{r}$

W = (6 i + 4 j - 2 k) . ( 2.5 i - 5.5 j + 1.5 k)

W = 15 - 22 - 3 = - 10 J

4 0

3 years ago

Which process creates x-rays

Zanzabum

Where are the answers?

5 0

4 years ago

Let’s say you have a cart of some mass and when pushed with 10N of force, the cart accelerates at 5.0 m/s/s. If you were to push

Korolek [52]

Answer:

a. The acceleration would increase.

Explanation:

because we know that

F=ma

m= mass and a= acceleration

so Mass is same for the cart in any situation that's why only acceleration could increase.

7 0

3 years ago

Read 2 more answers

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago