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2 years ago

Can a golf ball bounce like a rubber ball? Due in 1 hour. 15 points.

Physics

1 answer:

klasskru [66]2 years ago

5 0

Answer:

So, I tested it out and came to a conclusion.

The rubber ball is quite bouncy due to the fact that it is made of rubber which is what makes it bounce back and high, but although the golfball does in fact bounce back it does not bounce as high as the rubber ball due to the fact that it is made of plastic.

Explanation:

Hope this helps you out!

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Light of wavelength 505 nm passes through a single slit of width 4.32 x 10-5 m. At what angle does the first interference minimu

Nataly_w [17]

Answer:

0.665

Explanation:

I did the work. Just plug everything in from the formula. Look at the lesson manual.

3 0

3 years ago

How do you calculate a bearing angle and its equivalent angle?

denis-greek [22]

Explanation:

A bearing if an angle is measured clockwise from north direction.

e.g Below the bearing of B from A is 025. (3 figures are always given). the bearing of A from B is 205°.

4 0

2 years ago

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

GaryK [48]

The boat's position $x$ relative to its starting point $x_0=0$ is determined by

$x=x_0+v_0t+\dfrac12at^2$

where $v_0$ is its initial velocity, $a$ is its acceleration, and $t$ is time. After $t=6\,\mathrm s$ , the boat has traveled

$x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2$

$\implies x=91\,\mathrm m$

3 0

3 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

Instantaneous angular speed is: Group of answer choices the rate at which the angular acceleration is changing total angular dis

Ludmilka [50]

Answer: magnitude of the instantaneous angular velocity

Explanation:

Instantaneous angular speed is refered to as the magnitude of the instantaneous angular velocity. We should note that the instantaneous angular velocity is the rate that has to do with the rotation of an object in circular path.

5 0

3 years ago