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2 years ago

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A company wants to donate time and money into abuse prevention and wants to work with a government agency on the project. Which

federal agency should it work with?
Occupational Safety and Health Administration

State Board of Health

US Department of Health and Human Services

Centers for Disease Control and Prevention

2 answers:

pav-90 [236]2 years ago

7 0

US Department of Health and Human Services

GarryVolchara [31]2 years ago

4 0

<span>US Department of Health and Human Services</span>

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An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje

Delvig [45]

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

8 0

2 years ago

Mary takes 6.0 seconds to run up a flight of stairs that is 102 meters long. if mary's weight is 87 newtons, what power has mary

pentagon [3]

Thank you for posting your question here at brainly. I hope the answer will help. Below are the choices that can be found elsewhere:

<span>A. 1.5 * 10^3 Watts
B. 7.3 * 10^2 Watts
C. 3.5 * 10^2 Watts
D. 2.5 * 10^2 Watts
</span>
<span>Work = force*displacement = 10^2*87 = 8,700 joule
Power = work/time = 8,700/6 = 1.45*10^3 (rounded up to 1.5 kw). The answer is A. </span>

3 0

2 years ago

Read 2 more answers

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

2 years ago

Read 2 more answers

A hot–air balloon is moving at a speed of 10 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–dir

IrinaVladis [17]

The ball has an initial speed of 10m/s. This is because it is moving with the balloon. Now the balloonist throws the ball 4m/s with respect to himself, so it means that he gives the ball a extra push of 4m/s, so the total speed is 14m/s. Since it takes 30 seconds to reach the ground, the distance travelled is 14*30=420m.

7 0

2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

2 years ago