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Sveta_85 [38]
3 years ago
5

Hello:)! I don’t really understand the uniform deceleration too :/

Physics
1 answer:
NARA [144]3 years ago
4 0

Answer:

Deceleration is the opposite of acceleration.

The mathematical formula for deceleration is:

Deceleration = \frac{Final velocity - Initial velocity}{Time taken}

Uniform deceleration of an object can be defined as  decreasing the velocity of an object in equal amount in an equal interval of time.

The graph attached below shows a constant decrease in the velocity with a uniform time interval.

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At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0
3 years ago
A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc
guajiro [1.7K]

Answer:

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

a_o=a\times e^{-kt}

where,

k = rate constant  

t = age of sample

a_o = let initial amount of the reactant  

a = amount left after decay process  

We have :

a_o=x

a=58\%\times x=0.58x

t = 95 s

0.58x=x\times e^{-k\times 95 s}

\k= 0.005734 s^{-1}

Half life is given by for first order kinetics::

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{0.005734 s^{-1}}=120.86 s

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0
3 years ago
A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam
ahrayia [7]

Answer:

  1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

  (1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

8 0
3 years ago
Read 2 more answers
Which is the BEST description of how eyeglasses work?
Sonbull [250]

Answer:

C I think.

Explanation:

3 0
3 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
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