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liq [111]
3 years ago
6

I need HELP Please !!

Physics
1 answer:
aliya0001 [1]3 years ago
5 0
You have to figure it out
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What likely happens to the hart rate as the cardiac mucle gets stronger
Kaylis [27]
The heart rate will likely decrease. As the cardiac muscle, or heart, gets stronger, it takes less effort to pump more blood. As a result, the heart will probably beat less, decreasing the heart rate. This is why athletes often have lower heart rates than the average person.
5 0
3 years ago
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:
zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

5 0
1 year ago
3. During a tug-of-war, Team A pulls with a
Neko [114]

Answer:

8000 - 5000 =3000

Explanation:

8 0
3 years ago
Read 2 more answers
Blythe and Geoff are ice-skating together. Blythe has a mass of 40 kg and Geoff has a mass of 79 kg. Blythe pushes Geoff in the
Deffense [45]
9.9 m/s

calculations below 

4 0
3 years ago
Read 2 more answers
With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.
viktelen [127]

Answer:

0.976 c

Explanation:

v_{1e} = velocity of object 1 relative to earth = 0.80 c

v_{21} = velocity of object 2 relative to object 1 = 0.80 c

v_{2e} = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}

v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}

v_{2e} = 0.976 c

6 0
3 years ago
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