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Fantom [35]
2 years ago
5

Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at 8 m/s from 50

m above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths 25 m above the ground? The starting height of the ball thrown upward is 1.5 m above the ground. Ignore the effects of air resistance. whats the answer in m/s?
Physics
1 answer:
BaLLatris [955]2 years ago
5 0

Answer:22.62 m/s

Explanation:

Given

two balls are separated by a distance of 50 m

Alex throws  the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity  exactly at the same time.

ball  from ground travels a distance of 25 m in t sec

For Person on tree  

25=ut+\frac{1}{2}gt^2

25=8t+\frac{1}{2}\times 9.81\times t^2--------1

For person at ground

23.5=ut-\frac{1}{2}gt^2---------2

Solve equation (1)

50=16t+9.81t^2

9.81t^2+16t-50=0

t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s

put the value of t in equation 2

23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}

u=\frac{35.744}{1.58}=22.62 m/s

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A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?
kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

                                                                          = 1092 ft³

The weight of the air = density  x  volume

                                   = 0.07 lbs/ft³  x  1092 ft³

                                   = 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0
2 years ago
Which term describes speed in a certain direction?
11Alexandr11 [23.1K]

Answer:

A. Velocity

Explanation:

Velocity is vector quantity thus has both magnitude and direction. It describes not only the speed but also the direction. Speed is scalar quantity so describes only speed but not direction. Energy has nothing to do with speed, acceleration describes change in velocity in a direction over time

8 0
2 years ago
Read 2 more answers
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

8 0
3 years ago
12. The diameter of a circle is 2.42m. Calculate its<br>area in proper significant figure​
Sever21 [200]

Answer:

A = 4.6 [m²]

Explanation:

The area of a circle can be calculated by means of the following equation.

A=\frac{\pi }{4} *D^{2}

where:

A = area [m²]

D = diameter = 2.42 [m]

Now replacing:

A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]

7 0
2 years ago
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im
Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0
3 years ago
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