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Assoli18 [71]
2 years ago
5

A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?

Physics
1 answer:
ladessa [460]2 years ago
7 0
<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
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When momentum is conserved it is called _____. (multiple choice)
Klio2033 [76]

Answer:

Based off the word "conserved" I would say

A. Conservation of Momentum.

Explanation:

8 0
3 years ago
What is the resistance (R) when voltage is 179V and current is 5 Amps?
Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

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in our case:

R = 179 V / 5 A = 35.8 Ω

3 0
2 years ago
A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is
dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

  • <em>Mass of the pitcher, m₁ = 50 kg</em>
  • <em>Mass of the baseball, m₂ = 0.15 kg</em>
  • <em>Velocity of the ball, u₂ = 35 m/s</em>

<em />

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0
2 years ago
What is the unit for height
juin [17]
The unit of height is:
Feet
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Centimeters 

5 0
2 years ago
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Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt
Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

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If I am  1 m away from B , the path difference will be

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So path difference becomes odd multiple of  λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

8 0
3 years ago
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