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Lerok [7]
3 years ago
15

The acceleration reaches its minimal value of zero at the top of the trajectory. *

Physics
1 answer:
Serhud [2]3 years ago
4 0

it is true. the trajectory reaches the value of zero at the top

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Which of the following illustrates an increase in potential energy? Group of answer choices a wind-up toy winding down a person
horrorfan [7]

Answer:

A person climbs a set of stairs

Explanation:

Potential energy is said to be possessed by an object due to its position. As the height from the ground level increase, the potential energy increases. It is calculated by the below formula as :

P = mgh

Out of the given options, the option that illustrates an increase in potential energy is option (b) i.e. a person climbs a set of stairs. As he steps one stair, its position from ground increases. It means its potential energy increases.

7 0
3 years ago
PLS ANSWER DUE LATER TODAY !!!
Assoli18 [71]
Magna Carta. I’m pretty sure it’s the correct answer.
4 0
3 years ago
Read 2 more answers
• La longitud de un alambre de un sujetador de papel (clip) extendido mide 86 mm. ¿Cuántos sujetadores se pueden obtener de un r
kari74 [83]

Answer:

34883.7

Explanation:

Primero hay que convertir las dos unidades a la misma unidad-cm-

3km- 300000 cm

86mm- 8.6mm

Después hay que dividir

300000/8.6= 34883.7209302 ~ 34883.7

7 0
3 years ago
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
The strength of the electric field of a charged particle becomes greater as the distance from the particle increases. T F
natali 33 [55]
The answer is T it is true
7 0
2 years ago
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