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ahrayia [7]
4 years ago
13

When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le

ngth of the shorter stick in feet?
Physics
1 answer:
77julia77 [94]4 years ago
7 0

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

x+2.93

Both cover a magnitude of 8.32 ft, therefore

x +(x+2.97) = 8.32

Now solving for x we have,

x + (x + 2.93) = 8.32

2x + 2.93 = 8.32

2x = 8.32 - 2.93

x = \frac{ 8.32 - 2.93}{2}

x = 2.695 ft

Therefore the shorter stick is 2.695ft long.

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Answer:

d.) provides proteins for plants​

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What is the defination o<br>f temperature​
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How would you distinguish a scientific theory from a scientific law?
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4 years ago
The radius of the earth is 6380 km and the height of mt.everest is 8848 m. if the value of of acceleration due to gravity on the
Bas_tet [7]

a) 9.80 m/s^2

The acceleration due to gravity at a certain location on Earth is given by

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

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

h is the altitude above the Earth's surface

At the top of Mt. Everest,

R = 6380 km = 6.38\cdot 10^6 m

h' = 8848 m

g'=9.77 m/s^2

With

g'=\frac{GM}{(R+h')^2} (1)

At the Earth's surface,

R = 6380 km = 6.38\cdot 10^6 m

h = 0

g = ?

So

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

By doing the ratio (2)/(1), we find an expression for g in terms of g':

\frac{g}{g'}=\frac{\frac{GM}{R^2}}{\frac{GM}{(R+h')^2}}=\frac{(R+h')^2}{R^2}=\frac{(6.38\cdot 10^6+8848)^2}{(6.38\cdot 10^6)^2}=1.003

And therefore,

g=1.009g'=1.009(9.77)=9.80 m/s^2

b) 519.3 N

The weight of an object near the Earth's surface is given by

W=mg

where

m is the mass of the object

g is the acceleration of gravity at the object's location

In this problem,

m = 50 kg is the mass of the object

g' = 9.77 m/s^2 is the acceleration of gravity on top of Mt Everest

Susbtituting,

W=(50)(9.77)=519.3 N

6 0
3 years ago
A ball is rolled off a horizontal tabletop so that it leaves the tabletop and takes 0.35 seconds to reach the ground. The ball l
Rzqust [24]

kinematics allows to find the kinetic variables and the relationships between position, speed and acceleration, for launching projectiles we can find:

a)  The horizontal velocity  vₓ = 2.1 m/s

b) The height of the table y₀= 0.60 m

given parameters

  • Time to reach the floor t = 0.35 s
  • Horizontal distance x = 0.75 m

to find

   a. Horizontal speed

   b. table height

Let's use the kinematics relations, let's start by defining a reference frame with the x-axis along the motion of the ball and the vertical y-axis with the positive direction up.

We look for the initial velocity in the x axis, there is no acceleration we can use the uniform kinematic relations

         vₓ = x / t

Where vx is the horizontal velocity, g the acceleration of gravity (g=9.8 m/s²) and t the time

         vₓ = 0.75 / 0.35

         vₓ = 2.14 m / s

We look for the height of the table that corresponds to the initial height (y₀) of the ball, when reaching the floor the height is zero, on the vertical axis the initial velocity is zero (v_{oy} = 0)

        y = y₀ + v_{oy} t - ½ g t²

        0 = y₀ + 0 - ½ g t²

        y₀ = ½ g t²

        y₀ = ½ 9.8 0.35²

        y₀i = 0.60 m

  • In conclusion with the kinematic relations for the launch of projectiles we can find the horizontal velocity vₓ = 2.1 m / s and the height of the table i = 0.60 m

learn more about projectile launch here:

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