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iris [78.8K]
3 years ago
13

A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista

nce can be ignored, how high does it go?\
Physics
1 answer:
solniwko [45]3 years ago
3 0
Convert the given in SI units.

         (44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec

The distance traveled and the initial velocity can be related through the equation,
  
   d = (Vf)² - (Vi)²/ 2a

where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,

    d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)

The value of d from the equation,

    d = 9.17 meters

Convert this to feet,

    d = (9.17 m)(3.28 ft / 1 m) = 30 ft

Answer: 30 ft
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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
2 years ago
A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh
mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

d = vt = \omega r\times  t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

4 0
3 years ago
A proton is placed in an electric field of intensity 700 n/
kakasveta [241]
F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s²  along the direction of the field line

8 0
2 years ago
Please help and I don't mean to sound rude but, ONLY ANSWER IF YOUR GOING TO DO ALL 4 QUESTIONS
anzhelika [568]

Answer:

for first question is 2

for second question 1

for third question 2

for forth question 1

Explanation:

i hope i helped

6 0
2 years ago
Nuclei of u-238 atoms are
Elis [28]
 <span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
5 0
3 years ago
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