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olga55 [171]
3 years ago
15

Lactic acid is produced during low-moderate exercise True False

Physics
2 answers:
patriot [66]3 years ago
7 0
True. 

       Hope this helps!!
Marizza181 [45]3 years ago
3 0
The answer to your question is true.
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A vector quantity must include both magnitude and direction. Which measurement is a vector quantity? A) the rain accumulation at
bogdanovich [222]

Answer:

B)The motion of water in an ocean current

Explanation:

With respect to measurements, a vector has both a magnitude and a direction. The first three examples (maximum height of a hill, air temperature, and rain accumulation) are magnitudes only. The fourth example (motion of water in an ocean current) is a vector, because it has a magnitude (speed) and a direction (with the current).

4 0
3 years ago
An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-f
IgorC [24]

Resistance = (voltage) / (current)

Resistance = (6.0 v) / (2.0 A)

Resistance = 3.0 ohms 
7 0
3 years ago
A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How
nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0
3 years ago
In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0
3 years ago
Read 2 more answers
A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.
denpristay [2]

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

8 0
3 years ago
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