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Vladimir [108]
4 years ago
5

Which of the following is NOT an effective pre-exercise energizing snack?

Physics
2 answers:
maks197457 [2]4 years ago
4 0
Cookies would not be an effective pre-exercise energizing snack, as their high sugar content would be absorbed quickly by the body, not sustaining the athlete for the duration of their event.
Rom4ik [11]4 years ago
4 0
The answer is cookies
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A patient with type A blood is accidentally given a transfusion of type AB blood. Which of the following correctly describes wha
Vladimir [108]
<span>The antibodies in the patient's blood will fight or attack the donor's red blood cells and kill them. This could cause side effects called a hemolytic reaction - wherein the phenomena causes damage to the kidneys, fever, shortness of breath, low blood pressure, and chills.</span>
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3 years ago
Which of the following best describes a longitudinal wave?
Misha Larkins [42]

Answer:

A part not for sureokay

6 0
3 years ago
A car has 10,800 kg m/s of momentum as it collides with a trash dumpster AT REST.
Delvig [45]

Answer:

10800 kg m/s

Explanation:

Since the dumpster was initially at rest, its momentum before collision equal to zero. So that the total momentum of the car and dumpster before collision is 10800 kg m/s.

The principle of conservation of linear momentum states that: in a system of colliding bodies, total momentum is conserved. Therefore, the total momentum of the car and dumpster after collision is 10800 kg m/s.

3 0
3 years ago
A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.
Blizzard [7]

Answer:

Approximately 1.6\times 10^{3}\; \rm N.

Explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum  will be equal to the impulse that is applied to her.

The momentum p of an object is equal to the product of its mass m and velocity v. That is: p = m \cdot v.

Let v(\text{before}) and v(\text{after}) represent the velocity of the woman before and after the landing. Let m represent the woman's mass.

  • The woman's momentum before the landing would be m \cdot v(\text{before}).
  • The woman's momentum after the landing would be m \cdot v(\text{after}).

Therefore, the change in this woman's momentum would be:

\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}.

On the other hand, impulse is equal to force multiplied by the duration of the force. Let F represent the average force on the woman. The impulse on her during the landing would be F \cdot t.

Apply the Impulse-Momentum Theorem.

  • Impulse: F\cdot t.
  • Change in momentum: m \cdot (v(\text{after})- v(\text{before})).

Impulse is equal to the change in momentum:

F \cdot t = m \cdot (v(\text{after})- v(\text{before})).

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, v(\text{after}) = 0\; \rm m \cdot s^{-1}.

\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}.

3 0
4 years ago
Doubling an object’s height will have what effect on its potential energy due to gravity?
Gekata [30.6K]

Potential energy due to gravity = Ep = mgh [symbols have their usual meaning

Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.

3 0
4 years ago
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