Is Fahrenheit an SI unit?

A crowbar with the length of 200 cm is used to lift the load of 600 Newton . If the distance between fulcrum and load is 0.75 .

Leviafan [203]

Answer:

a. Effort = 960 Newton

b. Mechanical advantage (M.A) = 0.625

c. Velocity ratio (V.R) = 1.67

Explanation:

Given the following data;

Load = 600 N

Length of crowbar = 200 cm

Length of load arm = 0.75 m

Conversion:

100 cm = 1 m

X cm = 0.75 m

Cross-multiplying, we have;

X = 0.75 * 100 = 75 cm

First of all, we would find the effort arm;

Effort arm = length of crow bar - length of load arm

Effort arm = 200 - 75

Effort arm = 125 cm

Next, we would determine the mechanical advantage (M.A) of the crow bar;

Substituting the values into the formula, we have;

M.A = 0.625

To find the effort of the crow bar;

Making "effort" the subject of formula, we have;

Effort = 960 Newton

Lastly, we would determine the velocity ratio (V.R);

V.R = 1.67

8 0

2 years ago

If I work out rotational energy to be 102.2J which equals kg.M/s^2, and I hadn't factored time into it, would that be Joules per

Marina CMI [18]

Answer:

0.057 joules is needed to create the total rotational energy each second.

Explanation:

The energy rate is the ratio of total energy to time, which coincides with the definition of power at constant rate:

$\dot W = \frac{\Delta E}{\Delta t}$

$\dot W = \frac{102.2\,J}{\left(30\,min\right)\cdot \left(60\,\frac{s}{min} \right)}$

$\dot W = 0.057\,\frac{J}{s}$

$\dot W = 0.057\,W$

0.057 joules is needed to create the total rotational energy each second.

7 0

3 years ago

Two car horns are sounded creating two sound waves with frequencies that differ by a factor of three. How does the speed of the

Andrews [41]

Answer:

Remains the same

Explanation:

The speed of waves of higher and lower frequency both will be same.

the speed of sound in a medium is constant and independent of it's frequency. Moreover, when the frequency changes wavelength changes accordingly, such that their product remains constant.

we know that

υ×λ = constant = velocity

υ= frequency

λ= wavelength.

7 0

3 years ago

A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k

Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

7 0

3 years ago

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

5 0

3 years ago