When the moon is directly opposite the sun in the sky, its phase is?

You drop a 30 g pebble down a well. You hear a splash 2.7 s later. Ignoring air resistance, how deep is the well? Assume g = 9.8

lys-0071 [83]

I will assume here that the well is sufficiently short so that the time the sound takes to come from the bottom of the well to our ear is negligible.

Since the pebble moves by uniformly accelerated motion, the distance it covers is given by
$S= \frac{1}{2}gt^2$
where
$g=9.81 m/s^2$ is the gravitational acceleration
$t=2.7 s$ is the time the pebble takes to reach the bottom of the well

Therefore, the depth of the well is
$S= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2 = 35.7 m \sim 36 m$
and the correct answer is B.

8 0

3 years ago

The object experiences two external forces: 60 N to the west and 130 N to the east. What is the net external force?

MArishka [77]

Answer: 70 N to the East

Explanation:

lets assume east is positive and west is negative, since they are in opposite directions the net external force = F1+F2

Net force = (-60) + 130

Net force = 70

or

Net force = 70 N in the east direction

4 0

3 years ago

Which statements describe properties of stars?

Ede4ka [16]

Statements 1, 3, and 5 are true.

(A, C, and E)

7 0

3 years ago

Read 2 more answers

in an experiment the following readings were observed volume of alcohol flowing per minute equals to 10 raise to power - 5 cube

tekilochka [14]

Answer:

The viscosity is 1.30 x 10^-3 deca poise.

Explanation:

Volume per minute, V = 10^-5 m^3

Volume per second, V = 1.67 x 10^-7 m^3

density, d = 800 kg/m^3

radius, r = 0.05 cm

Length, L = 0.5 m

Height, h = 60 cm

Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa

Use the formula of rate of flow

$V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 = 3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise$

6 0

3 years ago

I’M DESPERATE PLZ HELP!! 40 POINTS!!

Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, $F_k$ , required is given by the formula, F = m × a,

From which we have;

$F_k$ = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, $F_k$ = 0.0625 N

c. The coefficient of friction between the ice and puck, $\mu_k$ , is given from the equation for the kinetic friction force as follows;

$F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8$

$\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755$

The coefficient of friction between the ice and puck, $\mu_k$ ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force $F_k = 0.012755 \times 1 \times 9.8 = 0.125 \ N$

The acceleration, a = $F_k$ /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here, $F_k = 0.125 \ N$

The coefficient of friction $\mu_k$ ≈ 0.012755 is constant

5 0

3 years ago