All categories

2 years ago

The minute hand of a large clock is 0.50 m long. (A) Calculate its linear speed at its tip in meters per second. (0.000873 m/s)

1 answer:

7 0

The rotational speed of the minute hand is 360 degrees in one hour or:
ω = 2π rad/3600 sec
or
ω = 0.001745 rad/sec

To convert this to linear speed, we use the formula:
v = rω

Substituting the given values:
v = 0.5 (0.001745) = 0.000873 m/s

For the centripetal acceleration:
a = v^2/r
a = (0.000873)^2 / 0.5
a = 1.52 x 10^-6 m/s^2

You might be interested in

The energy transfer diagram shows energy transfer in an MP3 player. Useful energy is transferred away from the MP3 player by lig

OLEGan [10]

Answer:

heat and sound.

Explanation:

Though some would argue that the heat is not useful. I guess it depends on if your hands are cold.

4 0

2 years ago

It took a squirrel 0.50\,\text s0.50s0, point, 50, start text, s, end text to run 5.0\,\text m5.0m5, point, 0, start text, m, en

STALIN [3.7K]

Answer:

-5.0m/s

Explanation:

3 0

2 years ago

Read 2 more answers

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m 3, po

liberstina [14]

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity = -3.1 m/s, negative sign indicates it is pointing downward.

acceleration = - 3.7m/ $s^2$ . Negative means acceleration is towards center of planet Mars.

Time taken = 3 seconds

$v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s$

So jumping velocity of Julia = 8 m/s

6 0

2 years ago

Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿

Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0

2 years ago

Consider a 2100-kg car cruising at constant speed of 70 km/h. Now the car starts to pass another car by accelerating to 110 km/h

Liula [17]

Answer: 51841.5 Watts

Explanation: Using the kinematic equation for the final velocity for a constant acceleration we have:

Vf=Vi+a*t

replacing the values the results is

a=(Vf-Vi)/t= (30.55 m/s-19.44 m/s)/5s= 2.22 m/s^2

Remenber that to convert the speed in Km/h to m/s we have to multiplier by the factor 0.277.

Finally to calculate the increment of power get the final velocity we have to use Neton second law to determine the Force applied to the car.

F=m* a=2100 Kg* 2.22 m/s^2= 4666.2 N

Then increment power to accelerate is given by:

ΔPower= Force* Δ velocity= 4666.2 N* 11,11 m/s= 51841.5 Watts

6 0

2 years ago