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2 years ago

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As a youngster, you drive a nail in the trunk of a young tree that is 3 meters tall. The nail is about 1.5 meters from the groun

d. Fifteen years later, you return and discover that the tree has grown to a height of 30 meters. About how many meters above the ground is the nail?

1 answer:

Lera25 [3.4K]2 years ago

8 0

Answer:

15m

Explanation:

Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

$\frac{30m}{3m} =10 times$

the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m

X=(1.5m)(10)=15m

the nail is 15 meters above ground level

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(1 point) Which of the following statements are true?A.The equation Ax=b is referred to as a vector equation.B.If the augmented

Anvisha [2.4K]

Answer:

A. False

B. False

C. True

D. True

E. True

F. True

Explanation:

A. The equation Ax=b is referred to as a matrix equation and not vector equation.

B. If the augmented matrix [ A b ] has a pivot position in every row then equation Ax=b may or may not be consistent. It is inconsistent if [A b] has a pivot in the last column b and it is consistent if the matrix A has a pivot in every row.

C. In the product of Ax also called the dot product the first entry is a sum of products. For example the the product of Ax where A has [a11 a12 a13] in the first entry of each column and the corresponding entries in x are [x1 x2 x3] then the first entry in the product is the sum of products i.e. a11x1 + a12x2 +a13x3

D. If the columns of mxn matrix A span R^m, this states that every possible vector b in R^m is a linear combination of the columns which makes the equation consistent. So the equation Ax=b has at least one solution for each b in R^m.

E. It is stated that a vector equation x1a1 + x2a2 + x3a3 + ... + xnan = b has the same solution set as that of the linear system with augmented matrix [a1 a2 ... an b]. So the solution set of linear system whose augmented matrix is [a1 a2 a3 b] is the same as solution set of Ax=b if A=[a1 a2 a3] and b can be produced by linear combination of a1 a2 a3 iff the solution of linear system corresponding to [a1 a2 a3 b] takes place.

F. It is true because lets say b is a vector in R^m which is not in the span of the columns. b cannot be obtained for some x which belongs to R^m as b = Ax. So Ax=b is inconsistent for some b in R^m and has no solution.

8 0

1 year ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Darya [45]

Answer:

$y = -8.37 cm$

Explanation:

As we know that the equation of SHM is given as

$y = A cos(\omega t)$

here we know that

$\omega = \frac{2\pi}{T}$

here we have

$T = 0.66 s$

now we have

$\omega = \frac{2\pi}{0.66}$

$\omega = 3\pi$

now we have

$y = (8.8 cm) cos(3\pi t)$

now at t = 2.3 s we have

$y = (8.8 cm) cos(3\pi \times 2.3)$

$y = -8.37 cm$

6 0

1 year ago

Read 2 more answers

A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

6 0

1 year ago

A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict

Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

3 0

2 years ago

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0

2 years ago