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gavmur [86]
3 years ago
15

An iron ball is dropped from a flat platform (note: the initial velocity is zero). What is the velocity of the ball in m/s 1,6 s

after it is dropped? Use g = 10 m/s2. Do not include the units in your answer.
Physics
1 answer:
julsineya [31]3 years ago
8 0
I believe that the correct answer is V=16mls
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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

where:

\omega=6.28 rad/s is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

v=(6.28)(0.65)=4.1 m/s

Learn more about angular speed:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
4 years ago
A man takes 20 seconds to climb 5m up a ladder. He weighs 720N. Calculate the power he must deliver to do this.
Inessa05 [86]

Answer:

180 W

Explanation:

The work done by the man against gravity is equal to its gain in gravitational potential energy:

W=mg\Delta h

where

(mg) = 720 N is the weight of the man

\Delta h= 5 m is the change in height

Substituting,

W=(720)(5)=3600 J

The power he must deliver is given by

P=\frac{W}{t}

where

W = 3600 J

t = 20 s is the time taken

Substituting,

P=\frac{3600}{20}=180 W

3 0
4 years ago
An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of
stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

               a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2

Option D is the correct answer.                

3 0
4 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
The frequency of the given sound is 1.5khz then how many vibration it is completing in one second ?​
kompoz [17]

Answer:

1500 per second.

Explanation:

vibrations = 1.5 kilohertz

1.5×1000=1500

the answer is 1500 per second.

6 0
3 years ago
Read 2 more answers
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