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2 years ago

15

How to solve for time given distance and velocity

1 answer:

Virty [35]2 years ago

8 0

Answer:

Well, I think you're talking about kinematics, especially uniform rectilinear motion. We know that there is a specific equation for that:

S = Vt + S0

With S being the distance, V the velocity, t the time and S0 the initial distance (initial displacement).

From this you can calculate t, if that's what you want.

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The 1st Law of Thermodynamics is a statement about (A) Temperatures scales (B) Whether a process can proceed in a specific manne

weqwewe [10]

Answer:

Energy conservation.

Explanation:

The 1st Law of Thermodynamics is a statement about energy conservation. It states that $\Delta U=Q-W$ , which means that if we <u>substract the work W done</u> by the system to the <u>heat Q given</u> to the system we get the <u>change in the internal energy</u> $\Delta U$ , so any excess in energy given to the system appears as internal energy, stating that energy is conserved.

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3 years ago

Find the intensity of a 55 dB sound given I 0=10^-12W/m^2

MakcuM [25]

Answer:

3.16 × $10^{-7}$ W/ $m^{2}$

Explanation:

β(dB)=10 × $log_{10}(\frac{I}{I_{0} })$

$I_{0}$ = $10^{-12}$ W/ $m^{2}$

β=55 dB

Therefore plugging into the equation the values,

55=10 $log_{10}(\frac{I}{ [tex]10^{-12}$ })[/tex]

5.5= $log_{10}(\frac{I}{ [tex]10^{-12}$ })[/tex]

$10^{5.5}$ = $\frac{I}{10^{-12} }$

316227.76× $10^{-12}$ = I

I= 3.16 × $10^{-7}$ W/ $m^{2}$

5 0

2 years ago

A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri

Alla [95]

Answer:

$q=1\times10^{-8}C$

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

$F=\frac{KQq}{d^{2}}$

By substituting the values

$0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}$

$q=1\times10^{-8}C$

Thus, the charge on the ball bearing is $q=1\times10^{-8}C$

7 0

2 years ago

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

6 0

3 years ago

The James Webb Space Telescope will be the largest telescope ever put into space and will replace the Hubble Space Telescope. It

Nikolay [14]

Answer:

B : It will avoid the distortion from the atmosphere and give scientists more accurate data

Explanation:

5 0

2 years ago