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Katen [24]
3 years ago
13

How often should cardio exercises be performed? A. Never B. Three to five times a month C. Three to five times a week D. Three t

o five times a day
Physics
2 answers:
EleoNora [17]3 years ago
6 0

Answer:

C. Three to five time a week.

Explanation:

Kobotan [32]3 years ago
4 0
C- three to five times a week
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Before the positive psychology movement, psychology focused mainly on
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tysm

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A .140kg baseball traveling 35.0m/s strikes the catchers mit bringing the ball to rest, recoils backward 11.0cm what’s the avera
Juli2301 [7.4K]

Average force applied by the call on the glove = 780 N

Explanation:

mass of baseball=0.140 kg

initial velocity = Vi= 35 m/s

Final velocity=Vf= 0

distance traveled=11 cm= 0.11 m

using the kinematic equation Vf²= Vi²+ 2 a d

where a = acceleration

0²= 35²+ 2 a (0.11)

a=-5568.2 m/s²

Now force is given by F= ma

F= 0.140 (5568.2)

F=-780 N

The negative sign signifies that the force acts in the opposite direction.

so the average force= 780 N

6 0
3 years ago
A friend tells you that a rowboat is propelled forward by the force of its oars against the water. First explain whether the sta
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It is correct, the action is paddling, where you move the water backwards, and the reaction is the boat moving forwards.
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3 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
2 years ago
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