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Irina-Kira [14]
3 years ago
7

A tortoise travels 15 meters (m) west, then another 13 centimeters (cm) west. How many meters total has she walked?

Physics
1 answer:
DENIUS [597]3 years ago
7 0

Answer:

15.13m

Explanation:

change 15m to cm

1m=100cm

15m times 100=1500cm

add 1500+13 = 1513cm

in metres it will be 15.13m

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Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa
Katarina [22]

Answer: minimum speed of launch must be 7.45m/s

Explanation:

Given the following:

Height or distance (s) = 2.83m

The final velocity(Vf) at maximum height = 0

Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2

From the 3rd equation of motion:

V^2 = u^2 - 2gs

Where V = final velocity

u = initial velocity

Therefore, u = Vi

u = √Vf^2 - 2gs

u = √0^2 - 2(-9.8)(2.83)

u = √0 + 55.468

u = √55.468

u = 7.4476 m/s

u = 7.45m/s

7 0
3 years ago
A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th
algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

v^2=u^2+2as

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 m/s^{2}

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as  

v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second

4 0
3 years ago
Select all that apply. There MIGHT be more than one.
Stolb23 [73]
I believe the answer to your question is “Lithosphere plate boundaries”

The planet Earth is covered by a layer formed by land and rocks called the earth's crust or lithosphere. This crust is not smooth and uniform, but rather irregular and composed of tectonic plates, also called lithosphere plates. These plates are not fixed as they are under the magma (high temperature molten rock).

Hope this helps!:)
4 0
2 years ago
Calculate the gravitational potential energy is a bicycle of mass 20kg and its rider of mass 50kg cycle to the top of a hill 120
Travka [436]

Explanation:

Gravitational Potential Energy can be calculated with the following formula:

mgh

Where m is mass, g is Gravitational Field Strength, and h is height. GFS on Earth is always 9.81, the combined mass of the cyclist and the bicycle is 70, and the height is 120. Multiplying these values together, we get:

82,404J.

7 0
11 months ago
On the surface of the earth the weight of an object is 200 lb. Determine the height of the
siniylev [52]

Answer:

The height of the  object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of Gmm_{e}

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

200=\dfrac{Gmm_{e}}{(3958.756)^2}

200\times(3958.756)^2=Gmm_{e}

Gmm_{e}=3.134\times10^{9}

We need to calculate the height of the  object

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

125=\dfrac{200\times(3958.756)^2}{r^2}

r^2=\dfrac{200\times(3958.756)^2}{125}

r^2=25074798.5

r=\sqrt{25074798.5}

r=5007.4\ miles

Hence. The height of the  object is 5007.4 miles.

7 0
3 years ago
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