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What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?

Igoryamba

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

5 0

3 years ago

Read 2 more answers

At what angle do the cars move after the collision?

liubo4ka [24]

Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

6 0

2 years ago

find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is

Readme [11.4K]

Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

7 0

3 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i

Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 $\frac{m}{s}$

Explanation:

The local ocean current is = 1.52 m/s

Direction $\theta$ = 40°

Velocity component in X - direction $V_{x}$ = 1.52 $\cos 40$ °

$V_{x}$ = 1.164 $\frac{m}{s}$

Velocity component in Y - direction $V_{y}$ = 8 + 1.52 $\sin 40$ °

$V_{y}$ = 8.97 $\frac{m}{s}$

The velocity of the ship relative to the earth

$V = \sqrt{V_{x}^{2} + V_{y} ^{2} }$

Put the values of $V_{x}$ and $V_{y}$ we get,

⇒ $V = \sqrt{1.164^{2} + 8.97 ^{2} }$

⇒ V = 9.05 $\frac{m}{s}$

This is the velocity of the ship relative to the earth.

7 0

4 years ago

If you wanted the pitch of a horn to drop relative to an observer, which way would you move the horn, relative to where that obs

Vladimir [108]

We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.

This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.

When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength.

Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.

From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>

3 0

3 years ago