A. Use the graph and the element made in question 2 to determine the mass of the star.

Check the correctness of formula t=2π √m/k, dimensionally

pantera1 [17]

Hi there! Lets see!

m is mass, and its units are kg
k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²

Therefore:

$\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s$

The period is given in seconds so the formula is dimensionally correct.

4 0

3 years ago

Half Time At 3:00 the hour hand and the minute hand of a clock point in directions that are 90.0

jenyasd209 [6]

Answer:

3 : 08 : 10.9

Explanation:

assuming a 12 hour clock

angular velocity of the hour hand is 2π/(3600(12)) rad/s

angular velocity of the minute hand is 2π/3600 rad/s

difference is 2π/3600 - 2π/(3600(12)) = 11(2π/(3600(12)) rad/s

45° = π/4 radians

This angle is covered in a time of

π/4 rad / 22π / (3600(12)) = 900(12) / 22 = 490.909090... s

or about 8 minutes 10.9 s

ANSWER 3:08:10.9

5 0

3 years ago

A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric fiel

jeka94

Answer:

$B=2.74\times 10^{-10}\ T$

Explanation:

It is given that,

A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m

We need to find the magnetic vector of the wave at the point P at that instant.

The relation between electric field and magnetic field is given by :

$c=\dfrac{E}{B}$

c is speed of light

B is magnetic field

$B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T$

So, the magnetic vector at point P at that instant is $2.74\times 10^{-10}\ T$ .

3 0

3 years ago

which is warmer -30 degrees Celcius or -30 degrees Fahrenheit? And how would i show my work for this question?

Alenkasestr [34]

Convert the temperature scales:

-> TC (Celsius Temperature)
-> TF (Fahrenheit Temperature)

->TC/5 = TF - 32/9

-30/5 = TF - 32/9

-270 = 5TF - 160

5TF = -270 + 160

5TF = -110

TF = -110/5

TF = -22 Fahrenheit

- 30 °C = -22 °F

So, -30°C is warmer than -30°F.

-30°F is lower than (30°C) -22°F.

4 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