All categories

3 years ago

A soccer ball takes 20 s to roll 10 m. What is the average speed of the soccer ball.

Physics

1 answer:

Georgia [21]3 years ago

8 0

Explanation:

velocity is distance divided by time.

the average speed of the ball is 10m/20s

= 0.5 m/s

You might be interested in

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

3 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers

Which statement explains how planets move in orbit as supported by Newton’s first law of motion?

GuDViN [60]

Answer:

Explanation:

got a one hundred on the test

5 0

3 years ago

Read 2 more answers

A nurse pushes a cart by exerting a force on the handle at a downward angle of 36 degrees below the horizontal. the loaded cart

Troyanec [42]

Since it was stated that it must move at constant velocity, so the only force it must overpower is the frictional force.

So the equation is:

F cos θ = Ff

F cos 36 = 65 N

F = 80.34 N

So the nurse must exert 80.34 N of force

4 0

3 years ago

A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t

Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

$Ff = m*a$ where Ff = μ * N and $a = \frac{V^2}{r}=\omega^2*r$

N - m*g = 0 So, N = m*g. Replacing everything on the original equation:

$\mu*m*g = m*\omega^2*r$ (eq2)

Solving for r:

$r = \frac{\mu*g}{\omega^2}=1.41m$

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0

3 years ago