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3 years ago

8

High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46

.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.

1 answer:

MissTica3 years ago

6 0

Answer:

The speed of the golf ball just after the impact is 73.04 m/s.

Explanation:

Given that,

The mass of golf club, m₁ = 183 g = 0.183 kg

The mass of golf ball, m₂ = 46.6 g = 0.0466 kg

The initial speed of golf club, u₁ = 58.6 m/s

The initial speed of a golf ball, u₂ = 0

The final speeds of club, v₁ = 40 m/s

We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s$

So, the speed of the golf ball just after the impact is 73.04 m/s.

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3 years ago

An electron moving with a velocity v= 5.0 × 107 m/s i enters a region of space where perpendicular electric and a magnetic field

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Answer:

The magnetic field that will allow the electron to go through the region without being deflected is $2\times 10^{-4}\ T$ .

Explanation:

Given that,

Velocity of the electron, $v=5\times 10^7\ m/s$

Electric field, $E=10^4\ V/m\ j$

We need to find the magnetic field that will allow the electron to go through the region without being deflected. It can be calculated as :

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3 years ago

A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r$

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$

So, the final E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

The magnitude of the E-field is

E = 2.66 N/m

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3 years ago

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o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

$\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}$

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1 year ago

if a moving object travels north for a distance of 105 m in 22 sec, what is it’s speed and velocity ?

artcher [175]

Answer:

Speed: 4.8 m/s

Velocity: 4.8 m/s north

Explanation:

Definitions:

- Speed is a scalar quantity, which is equal to the ratio between distance covered (d) and time taken (t):

$s=\frac{d}{t}$

- Velocity is a vector quantity, whose magnitude is equal to the ratio between the displacement of the object and the time taken:

$v=\frac{disp.}{t}$

And it also has a direction (the same as the displacement).

In this problem:

- The object travels a distance of

d = 105 m

In a time interval of

t = 22 s

So its speed is

$s=\frac{105}{22}=4.8 m/s$

- The displacement of the object is the same as the distance in this case, so still 105 m, covered in a time interval of 22 s; this means that the magnitude of the velocity is the same as the speed:

$v=4.8 m/s$

However, velocity is a vector quantity, so it also has a direction: and since the object has moved north, the direction of the velocity is north as well.

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3 years ago