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Aleks04 [339]
3 years ago
14

How far will a body move in 4 seconds if uniformly accelerated from rest at rate of 2m\sm\s​

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

16 m

Explanation:

For an object in uniformly accelerated motion (=with constant acceleration), the distance covered by the body can be found by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the body in this problem,

u = 0 (it starts from rest)

a=2 m/s^2 is the acceleration

Substituting t = 4 s, we find the distance covered:

s=0+\frac{1}{2}(2)(4)^2=16 m

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.<br> What happens to the average kinetic energy of water molecules as water freezes?
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because science

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3 years ago
The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d
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Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

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Angle =270^{\circ}

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

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angle turned =90^{\circ}

(f)Hour after that

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7 0
3 years ago
A first order reaction, A -&gt; products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

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0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

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

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
3 years ago
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