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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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The four lines observed in the visible emission spectrum of hydrogen tell us that

Gemiola [76]

Answer:

The correct answer is violet, blue, green, and red emission lines.

Explanation:

When samples of pure elements are heated they emit a continuous spectrum of electromagnetic radiation. When elements are heated at very high temperatures, the electrons get excited and they jump to consequent orbits and this results in transmission of electromagnetic radiation. The four lines visible in the emission spectrum of hydrogen are violet, blue, green, and red, the most intense of which is at 656 nanometre.

7 0

3 years ago

A proton moves through a region containing a uniform electric field given by E with arrow = 54.0 ĵ V/m and a uniform magnetic fi

madreJ [45]

Answer:

acceleration is 9.58 × $10^{7}$ (- 14 ĵ + 51 k ) m/s

Explanation:

given data

uniform electric field E = 54.0 ĵ V/m

uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T

velocity v = 170 î m/s.

to find out

acceleration

solution

we know magnetic force for proton is

i.e = e (velocity × uniform magnetic field)

magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )

magnetic force = e (- 68 ĵ + 51 k ) ..................1

and now for electric force for proton i.e

= uniform electric field × e ĵ

electric force = e (54 ĵ ) ............2

so net force will be add magnetic force + electric force

from equation 1 and 2

e (- 68 ĵ + 51 k ) + e (54 ĵ )

e (- 14 ĵ + 51 k )

so the acceleration (a) for proton will be

net force = mass × acceleration

a = e (- 14 ĵ + 51 k ) / 1.6 $10^{-19}$ / 1.67 × $10^{-27}$

acceleration = 9.58 × $10^{7}$ (- 14 ĵ + 51 k ) m/s

7 0

3 years ago

A 3kg box is pulled along the floor with a force of P=20 N at an angle of 30 degrees. If the box is moving at a constant velocit

Deffense [45]

If force is applied at 30 degree with the horizontal

then first we will find the components of force in two directions

$F_x = 20 cos20$

$F_y = 20 sin30$

now we can find normal force by balancing the forces in Y direction

$F_n + Fy = mg$

$F_n + 20 sin30 = 3*10$

$F_n + 10 = 30$

$F_n = 20 N$

now since block is moving with constant speed

so here forces along horizontal plane will be balanced also

$F_x = F_f$

$F_f = 20 cos30$

$F_f = 17.3 N$

Now the weight of the box is defined as

$W = mg$

now plug in all values

$W = 3*10 = 30 N$

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3 years ago

Determine the ratio β = v/c for each of the following.

nlexa [21]

Answer:

a) $\beta = 1.111\times 10^{-7}$ , b) $\beta = 9\times 10^{-7}$ , c) $\beta = 3.087\times 10^{-6}$ , d) $\beta = 2.5\times 10^{-5}$ , e) $\beta = 0.5$ , f) $\beta = 0.877$

Explanation:

From relativist physics we know that $c$ is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) <em>A car traveling 120 kilometers per hour</em>:

At first we convert the car speed into meters per second:

$v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 33.333\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 33.333\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 1.111\times 10^{-7}$

b) <em>A commercial jet airliner traveling 270 meters per second</em>:

The ratio $\beta$ is now calculated: ( $v = 270\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 9\times 10^{-7}$

c) <em>A supersonic airplane traveling Mach 2.7</em>:

At first we get the speed of the supersonic airplane from Mach's formula:

$v = Ma\cdot v_{s}$

Where:

$Ma$ - Mach number, dimensionless.

$v_{s}$ - Speed of sound in air, measured in meters per second.

If we know that $Ma = 2.7$ and $v_{s} = 343\,\frac{m}{s}$ , then the speed of the supersonic airplane is:

$v = 2.7\cdot \left(343\,\frac{m}{s} \right)$

$v = 926.1\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 926.1\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 3.087\times 10^{-6}$

d) <em>The space shuttle, travelling 27000 kilometers per hour</em>:

At first we convert the space shuttle speed into meters per second:

$v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 7500\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 7500\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 2.5\times 10^{-5}$

e) <em>An electron traveling 30 centimeters in 2 nanoseconds</em>:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

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

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 0.3\,m$ and $t = 2\times 10^{-9}\,s$ , then speed of the electron is:

$v = \frac{0.3\,m}{2\times 10^{-9}\,s}$

$v = 1.50\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 1.5\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.5$

f) <em>A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds</em>:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

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

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 10^{-14}\,m$ and $t = 0.38\times 10^{-22}\,s$ , then speed of the electron is:

$v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}$

$v = 2.632\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 2.632\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.877$

4 0

3 years ago

Alex Ar [27]

Answer:

please I do not understand that question

6 0

3 years ago