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Dimas [21]
3 years ago
12

5) (64(6-2) = A) 62 B) 8-9 C) 122 12-8

Physics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

B 8-9 because you had to subtract that number or simplify. then your answer is 8-9

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morpeh [17]
Yes it does that’s correct
6 0
3 years ago
A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc
Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity v_1  in time t_1

\frac{d}{2}=v_1t_1----1

from B to C

distance traveled is 0.5 d with v_2  and v_3  velocity for half-half time

\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2

divide 1 and 2 we get

\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}

\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}

Now average velocity is given by

v_{avg}=\frac{d}{t_1+t_2}

taking t_1  common

v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}

v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}

v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}  

6 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
PLEASE HELP!!! I have so much work left to do. PLEASE!!!
Neko [114]

Answer:

the answer is C.

Explanation:

6 0
2 years ago
3. A fighter jet covers a distance of 895m while accelerating from 22m/s to 35m/s. How long (in
larisa86 [58]

Answer:

31.404 seconds

Explanation:

To answer this equation, SUVAT is your best option utilizing and rearranging the known values to solve for the unknown.

here we have the values for

s=895

u=22

v=35

t= the unknown value

in this instant the equation s=0.5 x (u+v)t is the best equation to use

so we sub in the known values

895=0.5 x (22+35)t

rearrange to solve for t

895=28.5t

895/28.5=t

t=31.404 seconds (rounded to 3 decimal places)

6 0
2 years ago
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