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scoray [572]
3 years ago
9

A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. What is the maximum height reached by the basketball from its release point?
Physics
1 answer:
Hatshy [7]3 years ago
4 0

Answer:

The last one

Explanation:

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Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?
Helen [10]

Answer:

A. Mass and acceleration

Explanation:

  • According to Newton's second law of motion, the resultant force is directly proportional to the rate of change in momentum
  • Therefore; F = ma , where F is the resultant force, m is the mass, and a is the acceleration of the body.
  • <u>Resultant force depends on the acceleration and the mass of a body in motion, an increase in acceleration causes a corresponding increase in resultant force.</u> A body with higher mass will have a larger strong force if the acceleration is kept constant.
3 0
2 years ago
Read 2 more answers
What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?
Elodia [21]

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

8 0
2 years ago
Read 2 more answers
A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of
grigory [225]

Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

F_r=ma

Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

7 0
3 years ago
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
2 years ago
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
Veseljchak [2.6K]

To solve this problem we will apply the concepts related to the principle of destructive and constructive interference. Mathematically this expression can be given as

2nt = m\lambda

Here,

n = Index of refraction

t = Thickness

m = Order of the reflection

\lambda = Wavelength

We have all of this values, therefore replacing,

2(1.28)t = (1)(600nm)

t = 233nm    

Therefore the thickness of the oil slick is 233nm

3 0
2 years ago
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