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3 years ago

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For each value below, enter the number correct to four decimal places. Suppose an arrow is shot upward on the moon with a veloci

ty of 52 m/s, then its height in meters after t t seconds is given by h ( t ) = 52 t − 0.83 t 2 h(t)=52t-0.83t2. Find the average velocity over the given time intervals.

1 answer:

kenny6666 [7]3 years ago

4 0

The time intervals are missing in the question.You can put any time in equation v(t) which I have solved

Answer:

$v(t)=52-1.66t$

Explanation:

Given data

Velocity=52 m/s

Height h(t)=52t-0.83t²

To find

Average velocity

Solution

As we know that velocity is first derivative of distance with respect to time

So

$v(t)=\frac{dh(t)}{dt}\\ v(t)=\frac{d}{dt}[52t-0.83t^{2} ]\\ v(t)=52-1.66t$

After one second the velocity is

$v(t)=52-1.66t\\at\\t=1seconds\\v(1)=52-1.66(1)\\v(1)=56.33m/s$

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A monument has a height of 348 ft, 8 in. Express this height in meters. Answer in units of m.

tensa zangetsu [6.8K]

Answer:

The height of mountain in meter will be 106.2732 m

Explanation:

We have given height of mountain = 348 ft,8 in

We know that 1 feet = 0.3048 meter

So 348 feet $=348\times 0.3048=106.07meter$

And we know that 1 inch = 0.0254 meter

So 8 inch $8\times 0.0254=0.2032m$

So the total height of mountain in meter = 106.07+0.2032 = 106.2732 m

The height of mountain in meter will be 106.2732 m

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3 years ago

a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th

ludmilkaskok [199]

Answer:

Position at t= 4 seconds is 144 m

Explanation:

It is given that acceleration, a = 18 t, where t is the time.

We know that Velocity, $v = \int { a} \, dt$

Substituting value of a,

Velocity, $v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c$

We know that at t = 0, v = -12 m/s

So, $9*0^2+c=-12\\ \\ c=-12m/s$

So velocity, $v = (9t^2-12)m/s$

We also know that displacement, $x = \int { v} \, dt$

Substituting value of v,

Displacement, $x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c$

We know that at t = 0, particle is at origin, x =0.

So, $0=3*0^3-12*0+c\\ \\ c=0$

Displacement, $x = 3t^3-12t$

At t = 4 seconds

$x = 3*4^3-12*4=192--48=144m$

Position at t= 4 seconds is 144 m

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3 years ago

What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N

VMariaS [17]

The gravitational force between Mars and the Sun is $1.65\cdot 10^{21} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the Sun

$m_2 = 6.39\cdot 10^{23} kg$ is the mass of Mars

$r=229\cdot 10^6 km = 229\cdot 10^9 m$ is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N$

So, the closest answer is

$1.65\cdot 10^{21} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

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In Rutherford's gold foil experiment, some of the positive particles would pass through the foil and some would bounce off. This led to a new theory that all of the positive subatomic particles were in the center of the atom instead of evenly spread throughout.

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