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sleet_krkn [62]
2 years ago
11

A soccer ball kicked with a force of 13.5 n accelerates at 6.5 m/s^2 to the right. what is the mass of the ball?

Physics
1 answer:
natita [175]2 years ago
6 0

Answer:

2.08 kg

Explanation:

Newton's second law states that the acceleration of an object is proportional to the force applied to the object, according to the equation:

F=ma

where F is the force applied, m is the mass of the object and a its acceleration.

In this situation, the soccer ball is kicked with a force F=13.5 N and its acceleration is a=6.5 m/s^2, therefore its mass is

m=\frac{F}{a}=\frac{13.5 N}{6.5 m/s^2}=2.08 kg

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Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
2 years ago
While it’s impossible to design a perpetual motion machine, that is, a machine that keeps moving forever, come up with ways to k
MissTica
A perpetual motion machine is (as the name implies) a machine that moves perpetually; it never stops. Ever. So if you created one today and set it going, it would keep on going until the Big Freeze<span>. Calling that “a long time” is an understatement of epic proportions</span>
7 0
2 years ago
Read 2 more answers
Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show
Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call R_M its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

R_M = 1000 R (1)

Both  the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

V_M = V_R

Using Ohm's law, V=RI, we can rewrite the previous equation as:

R_M I_M = R I_R

where

I_M is the current in the meter

I_R is the current in the bulb

Using (1), this equation becomes

(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}

so, the current in the meter is 1000 times less than through the bulb.

5 0
2 years ago
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo
padilas [110]

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

      d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

        tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

     d = λ  L / y

     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

      λ = 269.19 10⁻⁹ m

4 0
2 years ago
In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
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