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olya-2409 [2.1K]
3 years ago
7

The length of the minute hand of the clock is 14cm. Calculate the speed with which the tip of the minute hand moves

Physics
1 answer:
Molodets [167]3 years ago
8 0

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

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

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

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What does a animal and a plant cell have in common.
Allushta [10]

Answer:

Structures that are common to plant and animal cells are the cell membrane, nucleus, mitochondria, and vacuoles. Structures that are specific to plants are the cell wall and chloroplasts.

Explanation:

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2 years ago
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A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The
Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

3 0
3 years ago
What are two ways to change the state of a substance?
weeeeeb [17]
Solid to liquid
Liquid to solid

By adding or removing heat energy aka thermal energy 
7 0
3 years ago
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A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work)
frosja888 [35]

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}

Time taken by light = 2.8583×10⁻⁶ seconds

3 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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