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2 years ago

If the moon disapared what affect would this have on earths tides

1 answer:

6 0

If the moon disapared what affect would this have on earths tides:
There would no longer be any tides.
The moon is what causes the push and the pull of waves.

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A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___

34kurt

Answer:

The final displacement from the origin is 1.6 km to the NE

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8· $\hat j$ , in vector form

2) 1.0 km East = 1.0· $\hat i$ , in vector form

3) 1.6 km South = -1.6· $\hat j$ , in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8· $\hat j$ + 1.0· $\hat i$ +(-1.6· $\hat j$ ) = 1.2· $\hat j$ + 1.0· $\hat i$

Final displacement, = 1.0· $\hat i$ + 1.2· $\hat j$

Where;

Δx = The displacement in the x-direction = 1.0· $\hat i$

Δy = The displacement in the y-direction = 1.2· $\hat j$

The magnitude of the resultant displacement vector is given as follows

$\left | d \right |$ = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0

3 years ago

The basic building block of life is

snow_tiger [21]

The basic building block of life is molecules, more specifically macromolecules. There are four different macromolecules which could all be described as the building blocks of life, namely carbohydrates, proteins, nucleic acids and proteins.

8 0

3 years ago

An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m

Aneli [31]

Answer:

$\displaystyle y_m=3.65m$

Explanation:

Motion in The Plane

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of $v_o$ and $\theta\\$ as the initial speed and angle, then we have

$\displaystyle v_x=v_o\ cos\theta$

$\displaystyle v_y=v_o\ sin\theta-gt$

$\displaystyle x=v_o\ cos\theta\ t$

$\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}$

If we want to know the maximum height reached by the object, we find the value of t when $v_y$ becomes zero, because the object stops going up and starts going down

$\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt$

Solving for t

$\displaystyle t=\frac{v_o\ sin\theta }{g}$

Then we replace that value into y, to find the maximum height

$\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2$

Operating and simplifying

$\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}$

We have

$\displaystyle v_o=20\ m/s,\ \theta=25^o$

The maximum height is

$\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}$

$\displaystyle y_m=3.65m$

7 0

2 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

2 years ago

If the mass of a material is 49 grams and the volume of the material is 10 cm^3, what would the density of the material be?

Irina-Kira [14]

Answer: 4.9 g/cm^3

Explanation.

!) Data

mass = 49 g
Volume = 10 cm^3

2) Formula

Density = mass / volume

Density = 49 g / 10cm^3 = 4.9 g/cm^3

This is a problem where you only need to know the formula and subtitute the numbers to calculate.

The formula of density is the very same definition: mass per unit of volume.

So, as you had the mass and the volume you just need to plug the values in the formula density = mass / volume.

8 0

2 years ago