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Aleks04 [339]
3 years ago
12

A tennis ball moving at a speed of 4 m/s, collides with a bowling ball at rest. The tennis ball bounces back in the direction op

posite to the initial velocity, and the bowling ball moves very slowly too. Which object experiences the greater magnitude impulse during the collision? The bowling ball Neither, both experienced the same magnitude impulse It's impossible to tell none of these The tennis ball
Physics
1 answer:
Alex Ar [27]3 years ago
3 0

Answer:

Both experienced the same magnitude impulse

Explanation:

This is because, the impulse force is internal to the system of both the tennis ball and the bowling ball. It is an action-reaction pair. So, the force exerted on the tennis ball by the bowling ball equals in magnitude to the force exerted by the tennis ball on the bowling ball although, they are in opposite directions. This, both experienced the same magnitude impulse.

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A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?
Ad libitum [116K]

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

5 0
2 years ago
Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad
Xelga [282]
0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is  1.4129 x 10⁻⁸   the radius of the Earth.
6 0
3 years ago
Malia has developed Type II Diabetes. When she eats, her body
AlladinOne [14]

Answer:

The correct option is c.

Explanation:

Metabolism is a sum of anabolic and catabolic reactions. The body's inability to produce/synthesize enough insulin is the cause of type II diabetes. Generally, metabolism is the process in which most compounds (proteins, carbohydrates and lipids) are produced (anabolism) or broken down (catabolism) in the body. Insulin is a protein that can be produced in less amount due to metabolic disorder in the body.

Maria's disease means she already has an exponentially high amount of blood sugar against the required insulin to balance it out, hence the disease already slowed down her rate of metabolism (catabolism) of blood sugar EXCEPT she decides to increase of metabolism by medication and exercise.

3 0
3 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 maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
3 years ago
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