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Ket [755]
3 years ago
5

Do you think that individual health behaviors should be taken into consider for individuals whose healthcare costs are covered b

y social health programs like Medicare and Medicaid?
Physics
1 answer:
aivan3 [116]3 years ago
5 0

Answer:

no

Explanation:

because some people will not be able to afford it

You might be interested in
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0
3 years ago
My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s
marishachu [46]

Answer:

1) Periodically check the no stop or NDL time on their computers

2) The dive computer planning mode can be used if available

3) Make use of a dive planning app

4) Check data from the RDP table or an eRDPML

Explanation:

The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop

The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning

The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.

7 0
3 years ago
Nerve impulses in the human body travel at a speed of about 100 m/s. A 1.6 m tall man accidentally drops a hammer on his toe. Ho
EastWind [94]

time=distance/speed

1.6/100 secs = 0.016secs=16millisecs

4 0
3 years ago
A motorboat traveling with a current can go 160 km in 4 hours. against the current it takes 5 hours to go the same distance. Fin
MatroZZZ [7]
<h2>Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>

Explanation:

Let speed of motor boat be m and speed of current be c.

A motorboat traveling with a current can go 160 km in 4 hours.

   Distance = 160 km

   Time = 4 hours

    Speed = m + c

   We have

            Distance = Speed x Time

            160 = (m+c) x 4

            m + c = 40     --------------------- eqn 1

Against the current it takes 5 hours to go the same distance.

   Distance = 160 km

   Time = 5 hours

    Speed = m - c

   We have

            Distance = Speed x Time

            160 = (m-c) x 5

            m - c = 32     --------------------- eqn 2

eqn 1 + eqn 2

           2m = 40 + 32

             m = 36 km/hr

Substituting in eqn 1

               36 + c = 40

                      c = 4 km/hr

Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.

3 0
3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
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