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3 years ago

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A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds

of its travel. What is
its average speed during the time from t = 5 seconds to t = 10 seconds?
A.
2 meters/second
B.
3 meters/second
c. 4 meters/second
D. 5 meters/second
E. 6 meters/second

1 answer:

Nesterboy [21]3 years ago

7 0

Answer:

c. 4 meters/second

Explanation:

The formula to calculate average speed is:

$s = \frac{x_{2} - x_{1} }{ t_{2} - t_{1} }$

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A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea

Kitty [74]

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L² ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223 (-9)

so

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

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3 years ago

Which of the following is the correctly abbreviated SI unit describing the mass of an object

frosja888 [35]

Mass is Kg
Mol is the amount of substance
m is for meters

8 0

3 years ago

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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

b)

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

d)

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

e)

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

f)

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

4 years ago

A trout jumps, producing waves on the surface of a 0.8-mdeep mountain stream. If it is observed that the waves do not travel ups

alina1380 [7]

Answer:

The value is $v = 2.8 \ m/s$

Explanation:

From the question we are told that

The depth of the mountain is $d = 0.8 \ m$

Generally the velocity of the surface wave is mathematically represented as

$v = \sqrt{ g * d }$

=> $v = \sqrt{ 9.8 * 0.8 }$

=> $v = 2.8 \ m/s$

Generally using the Froude number is mathematically represented as

$Fr = \frac{ V }{ v }$

Here V is the velocity of the current

Given that the waves do not travel upstream, then the flow of the current is supercritical which means that

$\frac{ V }{ v } > 1$

=> $V > v$

=> $V > 2.8 \ m/s$

Hence the minimum velocity of the current is

$v = 2.8 \ m/s$

This because the velocity of the current is greater velocity of the surface wave , so minimum will be like the lowest possible value of V which is v

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3 years ago

What is a newton equal to in terms of units of mass and acceleration?

Dimas [21]

1 newton =
the force that accelerates 1 kilogram of mass at the rate of 1meter per sec²

1 N = 1 kg-m/s²

3 0

4 years ago

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