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Kay [80]
3 years ago
8

10. A worker uses a pulley system to raise a 24.0 kg carton 16.5 m. A force of 129 N is exerted and the rope is pulled 33.0 m. a

. What is the mechanical advantage of the pulley system?
Physics
1 answer:
lozanna [386]3 years ago
5 0

Answer: Machanical advantage of the machine is 1.86

Explanation: Machanical advantage of a machine is the ratio of the Force to overcome which is the load in this case 24kg * 10= 240N to the force exerted(Effort) to overcome the load in this case 129N.

So, we have

MA = load/effort

= 240N/129N

= 1.86.

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A sonar emits a sound signal of frequency 40000Hz, towards the bottom of the sea.
kap26 [50]

Answer:

0.000025s

Explanation:

Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s

8 0
3 years ago
An old-fashioned LP record rotates at 33 1/3 RPM
Ksenya-84 [330]

Answer:

Part a) \frac{5}{9}\ \frac{rev}{sec}

Part b) \frac{9}{5}\ \frac{sec}{rev}

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

33\frac{1}{3}\ \frac{rev}{min}

Convert mixed number to an improper fraction

33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}

Remember that

1\ min=60\ sec

Convert rev/min to rev/sec

\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}

Simplify

\frac{5}{9}\ \frac{rev}{sec}

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

\frac{9}{5}\ \frac{sec}{rev}

4 0
3 years ago
What is 0.0025 m^2 in cm^2? 1m = 100cm.
PSYCHO15rus [73]

Answer:

25 cm²

Explanation:

Meters and centimeters are both the units for measuring length. The SI unit of measuring length is meters.

Area is the quantity which measures the cross-section occupied by the object.

Thus,

Given that = Area = 0.0025 m²

To convert into cm²

1 m = 100 cm

So, 1 m² = 10000 cm²

So,

<u>Area = 0.0025 × 10000 cm² = 25 cm²</u>

6 0
3 years ago
Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17
Alexus [3.1K]

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

8 0
3 years ago
Your performance on the pacer is used to measure which components of health- related fitness?
77julia77 [94]
Your performance on the pacer test is used to measure cardiovascular components of health-related fitness
8 0
4 years ago
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