1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
harkovskaia [24]
3 years ago
12

a pulley of diameter 15.0 cm is driven by a motor that revolves at 10 rpm. the pulley drives a 2nd pulley with diameter 10.0 cm.

find the rpm of the 2nd pulley
Physics
1 answer:
suter [353]3 years ago
3 0
6.67prm

diam1/diam2 = rpm1/rpm2
You might be interested in
A vector has an x component of 25.0 units and a y
sammy [17]
Here is the answer to your question

8 0
3 years ago
Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with other metals to produce an allo
Novay_Z [31]

Answer:

Explanation:18kt alloy contains

i) 75% of gold

rhogold=19.3g/cm^3

=75/100×19.3

=14.475g/cm^3

ii) 16% of silver

rhosilver=10.5g/cm^3

=16/100×10.5

=1.68g/cm^3

iii) 9% of copper

rhocopper =8.90g/cm^3

=9/100×8.9

=0.801g/cm^3

Overall density of 18kt gold

=(0.801+1.68+14.475)g/cm^3

=16.956g/cm^3

=17g/cm^3 to 3s.f

6 0
3 years ago
A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave
Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction \mu =0.4

Consider a small element at an angle \theta having an angle of d\theta

Normal Force=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}

N=T\cdot d\theta

Friction f=\mu \times Normal\ Reaction

f=\mu \cdot N

and T+dT-T=f=\mu Td\theta

dT=\mu Td\theta

\frac{dT}{T}=\mu d\theta

\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta

\frac{T_2}{T_1}=e^{\mu \pi}

\frac{T_2}{T_1}=e^{0.4\times \pi }

\frac{T_2}{T_1}==e^{1.256}

\frac{T_2}{T_1}=3.51

7 0
3 years ago
(a) what will an object weigh on the moon's surface if it weighs 100 n on earth's surface
juin [17]
We know the equation

weight = mass × gravity

To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.

So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)

Therefore,

100N = mass × 10
mass= 100N/10
mass= 10 kg

Now, all we need are the moon's gravitational field strength and to apply this to the equation

weight = 10kg × (gravity on moon)
4 0
3 years ago
A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 17 feet. The ball is started in moti
Romashka-Z-Leto [24]

twelve because it is the answer

5 0
3 years ago
Read 2 more answers
Other questions:
  • A 11,000-watt radio station transmits at 880 kHz. Determine the number of joules transmitted per second.
    12·1 answer
  • A straight wire carrying a 3.30-A current is placed in a uniform magnetic field of magnitude 0.272 T directed perpendicular to t
    12·1 answer
  • PLEASE HELP TIME SENSITIVE:)
    10·2 answers
  • Eliza went for a swim. When she came out of the water, she felt cold. At the moment, a strong wind blew, and she felt even colde
    5·1 answer
  • If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
    9·1 answer
  • A transformer connected to a 120-V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is
    10·1 answer
  • 3.
    8·1 answer
  • How can you increase the current in a circuit?
    8·1 answer
  • A negative point charge - is at the center of a hollow insulating spherical shell, which has an inner radius R1 and an outer rad
    10·1 answer
  • What is the value of sin r and sin i in physics​
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!