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4 years ago

in positive numbers less than 1, the zeros between the decimal point and a non-zero number are _______ significant?

Physics

1 answer:

DedPeter [7]4 years ago

8 0

Answer:

Explanation:

If a number of less than 1, then the number has a decimal point like

0.085, 0.008 e.t.c.

The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.

But if there a zero between the none zero e.g. 0.0087056

Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6

But if the zero is not in between the none zero digit, then the zero is insignificant

E.g 0.05800

The last two zero is insignificant, the significant number is 5 and 8

So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.

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Two gliders collide on an air track moving in from opposite directions, and they bounce back. The masses of the gliders are 0.20

timurjin [86]

Answer:

so initial momentum is 0.22kgm/s

Explanation:

m1=0.20kg

m2=0.30kg

initial velocity of m1=u1=0.50m/s

initial velocity of m2=u2=0.40m/s

total momentum of the system before collision

Pi=m1u1+m2u2

Pi=0.20kg×0.50m/s+0.30kg×0.40m/s

Pi=0.1kgm/s+0.12kgm/s

Pi=0.22kgm/s

3 0

4 years ago

A car undergoing uniform acceleration travels 100 meters from a standing start in a given period of time. If the time were incre

iogann1982 [59]

Answer:

d = 1600 m

Explanation:

If car start from rest and accelerates uniformly then the displacement of the car is given as

$d = v_i t + \frac{1}{2}at^2$

now we know

$d = 0 + \frac{1}{2}at^2$

now if the same car will accelerate from rest for four times of the previous time interval

then we will have

$d = 0 + \frac{1}{2}a(4t)^2$

$d = 16(\frac{1}{2}at^2)$

so the distance will be 16 times more

$d = 1600 m$

5 0

4 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

4 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is <u>6.4m</u>

5 0

3 years ago

PLEASE HELP !!

LenKa [72]

Explanation:

It is given that

The kinetic energy of Mr. Thomson’s prius, K = 641300 J

He is moving with a velocity, v = 22 m/s

It is required to find the mass of Mr. Thomson’s prius.

Due to his motion, a kinetic energy will possess by him. The mathematical form of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2$

Solving for m

$m=\dfrac{2K}{v^2}$

Plugging all the values, we get :

$m=\dfrac{2K}{v^2}\\\\m=\dfrac{2\times 641300}{(22)^2}\\\\m=2650\ kg$

So, the mass of Mr. Thomson’s prius is 2650 kg.

8 0

4 years ago