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Naddik [55]
4 years ago
15

Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i

ndicate equal or different falling distances for successive seconds? 1. Greater distances fallen in successive seconds 2. All are wrong. 3. Smaller distances fallen in successive seconds 4. Equal distances fallen in successive seconds 5. Initially equal distances fallen in successive seconds, then greater distances fallen in successive seconds
Physics
1 answer:
maks197457 [2]4 years ago
3 0

Answer:

1  greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

       Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s)      Y (m)

  1          -4.9

  2         -19.6

   3        -43.2

The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

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A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %
olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground =  46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt²     ..............1

here d is distance and g is  acceleration and t is time

so when object falling it will be

h = 4.9 t²   ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)²       ...................3

so

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

we use here 3.771 s because  0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

so height is 69.68 m

6 0
4 years ago
A point charge is a model that represents a charge as acting uniformly on its surroundings. What other characteristic does a poi
inysia [295]

Explanation:

A point charge is a model that represents a charge as acting uniformly on its surroundings. The characteristics of a point charge are as follows :

  • Electric field lines are outwards if the charge is positive.
  • Electric field lines are inwards if the charge is negative.
  • If the distance from the point charge increase, the magnitude of the electric field decreases and vice versa.
3 0
3 years ago
A 55- kg woman has a momentum of 220 kg.m/s. What is her velocity?
exis [7]

Answer:

P=mv

Momentum=mass*velocity

Mass=55kg

Velocity=?

Momentum=220kg.m/s

P=mv

v=p/m

v=220/30

v=7.33m/s

Explanation:

6 0
3 years ago
Hi! How to do this? thank you in advance
devlian [24]

Answer: it says that the magnitude of average velocity is always equal to average speed so therefore it would be the last option

6 0
2 years ago
A 15.0-Ω resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed swit
Aloiza [94]

Answer:

0.04328 H

0.00288 seconds

0.01326 seconds

Explanation:

V = Voltage = 6.3 V

R = Resistance = 15 Ω

L = Inductance

I = Decayed current = 0.21 A

t = Time to decay = 2 ms

Maximum Current is given by

I_0=\frac{V}{R}\\\Rightarrow I_0=\frac{6.3}{15}\\\Rightarrow I_0=0.42\ A

Current in the circuit is given by

I=I_0exp\frac{-Rt}{L}\\\Rightarrow ln\frac{I_0}{I}=\frac{Rt}{L}\\\Rightarrow L=\frac{Rt}{ln\frac{I_0}{I}}\\\Rightarrow L=\frac{15\times 2\times 10^{-3}}{ln\frac{0.42}{0.21}}\\\Rightarrow L=0.04328\ H

The inductance is 0.04328 H

Time constant is given by

\tau=\frac{L}{R}\\\Rightarrow \tau=\frac{0.04328}{15}\\\Rightarrow \tau=0.00288\ s

Time constant is 0.00288 seconds

Time required is given by

t=\tau ln\frac{I_0}{I}\\\Rightarrow t=0.00288\times ln\frac{I_0}{0.01I_0}\\\Rightarrow t=0.01326\ s

The time to reach 1% of its original value is 0.01326 seconds

7 0
4 years ago
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