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3 years ago

11

you toss an apple from a tree 5m horizontally. The Apple hits your friend on the ground that is 7m away from the tree. How fast

did you throw the Apple in m/s? hello, answeerr

1 answer:

saw5 [17]3 years ago

8 0

-4.90 seconds -16.8 m/s-

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all of the following answer choices are ideas from daltons atomic theory. which one of them do we now know is not true?

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Did you forget to include the options?

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3 years ago

It is well known that runners run more slowly around a curved track than a straight one. One hypothesis to explain this is that

barxatty [35]

Answer:

$percentage = 6.27 %$

Explanation:

As we know that when runner is moving on straight track then the net force on his feet is given as

$F_s = mg$

while when runner is moving on circular track then we have

$F_c = \sqrt{(mg)^2 + (\frac{mv^2}{R})^2}$

$F_c = m\sqrt{g^2 + \frac{v^4}{R^2}}$

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now percentage change is given as

$percentage = \frac{F_c - F_s}{F_s} \times 100$

$percentage = \frac{m(10.42) - m(9.81)}{m(9.81)}\times 100$

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3 years ago

During summer in the northern hemisphere _____.

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For part of our orbit the northern half of Earth is tilted toward the Sun. This is summer in the northern hemisphere; there are longer periods of daylight, the Sun is higher in the sky, and the Sun's rays strike the surface more directly, giving us warmer temperatures

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3 years ago

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

svetlana [45]

Answer:

7 deg

Explanation:

$m$ = mass of the rod = $1.4 kg$

$W$ = weight of the rod = $mg = (1.4) (9.8) = 13.72 N$

$k_{L}$ = spring constant for left spring = $59 Nm^{-1}$

$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

Angle made with the horizontal is given as

$\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg$

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Answer:

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