What happens to light when it travels from air into water

Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

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

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

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

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image

lilavasa [31]

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor. T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0

3 years ago

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

3 years ago

4. A gas's solubility is best in a liquid solvent when the solution is under high or low pressure

Black_prince [1.1K]

Answer:

low pressure

Explanation:

6 0

3 years ago

Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci

Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation: Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F = m*a

∑ Fx = m* a(x) ∑ Fy = m* a(y)

También sabemos que el coeficiente de roce dinámico es:

μ = 0.2 = F(r)/N siendo N la fuerza normal.

Si descomponemos la fuerza P = mg = 20Kg* 9.8m/seg²

P = 196 [N] en sus componentes sobre los ejes x y y tenemos

Py = P* cos30 = 196* √3/2 = 98*√3

Px = P* sen30 = 196*1/2 = 98

La sumatoria sobre el eje y es :

∑ F(y) = m*a Py - N = 0 98*√3 = N ( no hay movimiento en la dirección y)

∑ F(x) = m*a P(x) - Fr = m*a

Fr = μ *N = 0.2* 98*√3

Fr = 19.6*√3 [N]

98 - 19.6*√3 = m*a

98 - 33.52 = m*a

a = (98 - 33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v² = v₀² + 2*a*d ( se pueden chequear unidades para ver la consistencia de la ecuación v y v₀ vienen dados en m/seg entonces v² y v₀² vienen en m²/seg², el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg² entonces es consistente la relación

v² = 0 + 2*3.22*80 ( la velocidad inicial es cero)

v² = 515.2 m²/seg²

v = √515.2 m/seg

v= 26.70 m/seg

6 0

3 years ago