30N of force is used to raise a box that weight 60N up an incline plane. What is the mechanical advantage of the incline plane?

A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the mot

nordsb [41]

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

6 0

3 years ago

What is the potential energy of an object 20 m in the air with a<br> mass of 600 kg?

Lana71 [14]

Answer:

Ep = 117600 J

Explanation:

Data:

Mass (m) = 600 kg
Height (h) = 20 m
Gravity (g) = 9.8 m/s²
Potential Energy (Ep) = ?

Use formula:

Ep = m * g * h

Replace:

Ep = 600 kg * 9.8 m/s² * 20 m

Multiply operations, and units:

Ep = 117600 J

What is the potential energy?

The potential energy is <u>117600 Joules.</u>

7 0

2 years ago

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and 4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and 1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and 2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and 2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge Right Charge Resulting force(N)

1μC 4μC 90 N

4μC 1μC 90 N

2μC 2μC 90 N

1μC 2μC 45 N

1μC 8μC 180 N

2μC 8μC 360 N

5 0

3 years ago

An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

4 0

3 years ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

schepotkina [342]

Answer:

6.77 minutes

Explanation:

172 degree - 78 degree = (185 degree - 78 degree)e−2 k

=> 94 = 107

e−2 k => 94 ÷ 107

k => ln (94÷107) / 2

147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]

=> 69 = 107 e^ [ln (94÷107) / 2]

e^[ln (94÷107) / 2] =69 ÷ 107

=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]

t=> -0.4387 ÷ -0.0648

t => 6.77 minutes.

Therefore, the final answer to the question is 6.77 minutes.

4 0

3 years ago