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3 years ago

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A ball is thrown upward with an initial velocity of +9.8 m/s. How high does it reach before it starts descending?

1 answer:

aksik [14]3 years ago

3 0

Hope this helps you.

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You decide to travel by car for your holiday visits this year. You leave early in the morning to avoid congestion on the roads.

podryga [215]

Answer:

a) d = 182.08 miles

b) $\overline{v} = 54.5 mph$

Explanation:

a) The distance can be found as follows:

$d_{T} = d_{1} + d_{2} + d_{3}$

$d_{T} = v_{1}*t_{1} + v_{2}*t_{2} + v_{3}*t_{3}$

$d_{T} = 67.4 mph*1.70 h + 0*23.4 min + 54.0 mph*1.25 h = 182.08 miles = 292.9 km$

b) The average speed can be calculated using the following equation:

$\overline{v} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}}$

Where "f" is for final and "i" for initial

$\overline{v} = \frac{182.08 miles - 0 miles}{(1.70 h + 23.4 min*\frac{1 h}{60 min} + 1.25 h) - 0 h} = 54.5 mph$

I hope it helps you!

5 0

3 years ago

(NO LINKS) Compare and contrast magnetism and electricity. And how are they alike? No need for a long answer.

lakkis [162]

Electricity and magnetism are essentially two aspects of the same thing, because a changing electric field creates a magnetic field, and a changing magnetic field creates an electric field

Explanation:

(Can I please have Brainiest? :)

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3 years ago

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CRITICAL THINKING HELP !!<br><br> How does predicting help your brain make connections ?

Zina [86]

It helps you understand the concept of what you are predicting

8 0

4 years ago

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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

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

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

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

$F_c = m\sqrt{9.8^2 + \frac{8^4}{18^2}}$

$F_c = m(10.42)$

now percentage change is given as

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

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

$percentage = 6.27 %$

6 0

3 years ago