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3 years ago

It's not important to monitor your heart rate during moderate intensity activities. True False

Physics

2 answers:

Mama L [17]3 years ago

5 0

I think it is False

hope this helps :3

8_murik_8 [283]3 years ago

5 0

Answer:

FALSE

Explanation:

Monitoring heart rate is important in all physical activities. Some heart problems are silent and may manifest during increased physical effort, albeit moderate. Therefore, the practice of exercises should be authorized by a cardiologist and you should monitor your heart rate whenever possible. This is a simple matter now, as there are applications and wristbands that measure heart rate.

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Suppose that you wanted to close a door by throwing a ball at the door. one ball will bounce off the door in a perfectly elastic

Wewaii [24]

The perfect elastic Collision ball will hit the door first because the clay is heavier and will take more time

5 0

3 years ago

as a hurricane or tropical storm approaches your location, it's forward speed decreases from 20 kt to 10 kt. How might this affe

zzz [600]

Answer:

The correct answer is c. More total rainfall from a slower moving storm.

When the the forward speed of the hurricanes and tropical storms slows down they tend to increase the rainfall. Because of the slow movement the storm can be for few days over a given region and produce rainfall without stopping, thus create major flooding, pilling up of the coastal water, and produce persistent strong winds even though they have decreased in their forward speed.

Explanation:

6 0

3 years ago

Read 2 more answers

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

SVETLANKA909090 [29]

Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is $\sqrt{a^2 + b^2}$

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

8 0

3 years ago

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0

3 years ago

Read 2 more answers

What is the volume of a rock with a density of 3.00 g/cm3 and a mass of 600g?

Mademuasel [1]

The equation of D = m/V

Where D = density
m = mass
and V = volume

We are solving for V, so with the manipulation of variables we multiply V on both sides giving us
V(D) = m
now we divide D on both sides giving us
V = m/D

We know our mass which is 600g and our density is 3.00 g/cm^3
so
V = 600g/3.00g/cm^3 = 200cm^3 or 200mL

a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL

If you wish to change it to L, you'd have to convert.

5 0

3 years ago