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Komok [63]
3 years ago
11

While riding on an elevator descending with a constant speed of 2.7 m/s , you accidentally drop a book from under your arm. How

long does it take for the book to reach the elevator floor, 1.0 m below your arm?
Physics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

0.45 s.

Explanation:

From Newton's equation of motion,

s = ut + 1/2gt²........................ Equation 1

Where s = distance, u = initial velocity, a = acceleration, t = time

Given: s = 1.0 m, u = 0 m/s (dropped from height), g = 9.81 m/s²

Substituting into equation 1,

1 = 0(t) + 1/29.81(t²)

1 = 9.81t²/2

2 = 9.81t²

t² = 2/9.81

t² = 0.204

t = √(0.204)

t = 0.45 s.

Hence it takes 0.45 s for the book to reach the elevator floor.

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What will be the average velocity of a body falling in free fall on Earth for 3 s?
SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time  = 3s

Unknown:

Average velocity  = ?

Solution:

To solve this problem, we use the expression below:

      v  = u + gt

v is the average velocity

u is the initial velocity  = 0m/s

g is the acceleration due to gravity  = 9.8m/s²

t is the time

So;

        v  = 0 + (9.8 x 3)  = 29.4m/s

6 0
2 years ago
PLEASE HELP SEE PART 2 FOR MORE INFO
Angelina_Jolie [31]

Answer:

This question cannot be answered

Explanation:

This is a practical experiment which can only be done in person. Kindly go through the instructions and do the experiment carefully.

8 0
2 years ago
Anybody got any answers???
dexar [7]

Answer:

b? to a?

Explanation:

6 0
3 years ago
Read 2 more answers
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
Why can't heat be transferred between two objects that have the same temperature
Inessa [10]

Answer: Because temperature is a measure of the average kinetic energy of the atoms or molecules in the system. The zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium; therefore, they are the same temperature.

Explanation:9 (- _ -)

3 0
3 years ago
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