Ask question

Ask question

All categories

3 years ago

7

Represent 8953 ms with Sl units having an appropriate prefix Express your answer to four significant figures and include the app

ropriate units. PÅ ROS? 8953 ms = 8.953 . 10 ms

1 answer:

tatyana61 [14]3 years ago

8 0

Answer:

8.953 s

Explanation:

Here a time of 8953 ms is given.

Some of the prefixes of the SI units are

mili = 10⁻³

micro = 10⁻⁶

nano = 10⁻⁹

kilo = 10³

Mega = 10⁶

Each prefix increases or decreases by a factor of 10³

The number is 8953.0

Here, the only solution where the number of significant figures is mili. If any other prefix is chosen then the significant figures will increase.

$1 milisecond = 1000 second$

$1\ second=\frac{1}{1000}\ milisecond$

$\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second$

So, 8953 ms = 8.953 s

You might be interested in

A star produces 2x10^26 watts. how much energy does it lose every minutes

Step2247 [10]

Answer:

Energy loss per minute will be $120\times 10^{26}j$

Explanation:

We have given the star produces power of $2\times 10^{26}W$

We know that 1 W = 1 J/sec

So $2\times 10^{26}W=2\times 10^{26}J/sec$

Given time = 1 minute = 60 sec

So the energy loss per minute $=2\times 10^{26}\times 60=120\times 10^{26}j$

We multiply with 60 we have to calculate energy loss per minute

7 0

3 years ago

You can see your image in a shiny, flat surface because the light waves are bouncing off the surface back at you. This is an exa

Alexxx [7]

I think it would be Reflection because the light in the ray reflects off of any flat or shiny object.... Hope this helps ^-^....

4 0

3 years ago

Read 2 more answers

Phenotype describes

vivado [14]

Answer is C...........

5 0

3 years ago

Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t

kykrilka [37]

Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

6 0

3 years ago