Answer:
The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds
Explanation:
The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend
The direction in which the student tosses the ball = The horizontal direction
Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0
The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;
h = 1/2 × g × t²
Where;
g = The acceleration due gravity of the ball = 9.81 m/s²
t = The time of motion to cover height, h
Then height is already given as h = 3.8 m
Substituting gives;
3.8 = 1/2 × 9.81 × t²
t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²
∴ t = √0.775 ≈ 0.88 seconds
The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.
If I am not wrong i thinks it is in the toroid uniforms
Answer:
Una Mezcla Homogénea es aquella mezcla en la que las sustancias que la forman poseen una combinación uniforme.Son ejemplos de Mezclas Homogéneas: Compuesta
Explanation:
Aire (es una mezcla de gases homogénea formada principalmente por de nitrógeno, oxígeno, vapor de agua, dióxido de carbono...)
Leche (mezcla de agua, carbohidratos, proteínas...)
Bebida alcohólica (mezcla de agua y alcohol etílico)
Acero (mezcla de elementos aleados como el hierro, el carbono y otras sustancias)
Petróleo (mezcla de hidrocarburos)
Agua de mar (mezcla de agua, cloruro sódico y otras sustancias)
Mezcla de agua y sal disuelta
Agua azucarada (mezcla de agua y azúcar)
Aleación metálica (las aleaciones metálicas son mezclas en las que se combinan diferentes metales de una manera homogénea y definida)
Perfume (mezcla de agua y otras sustancias olorosas cuya composición es uniforme)
Answer:
Explanation:
Volume per unit time flowing will be conserved
a₁v₁ = a₂ v₂
π r₁² x v₁ = π r₂² x v₂
(0.9 x 10⁻²)² x .35 = ( .45 x 10⁻² )² x v₂
v₂ = 1.4 m / s
Answer:
4.6 m
Explanation:
First of all, we can find the frequency of the wave in the string with the formula:
where we have
L = 2.00 m is the length of the string
T = 160.00 N is the tension
is the mass linear density
Solving the equation,
The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.
The n=1 mode (fundamental frequency) of an open-open tube is given by
where
v = 343 m/s is the speed of sound
Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:
