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3 years ago

What are the average velocity and average acceleration of the tip of a 2.4 cm long hour hand of a clock?

1 answer:

8 0

A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:

v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour

Therefore, the average velocity is 1.257 cm per hour.

For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.

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a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$

So,

Time in which train will come to rest = 20 seconds

4 0

2 years ago

Ideally, the resistance of an ammeter should be:

pav-90 [236]

Ideally the resistance should be ZERO

7 0

3 years ago

if a current of 0.3 ampere flows through a conductor of resistance of 11.7 ohm's,the voltage across the ends of the conductor is

Gnesinka [82]

Here,

Resistance = 11.7

Ampere = 0.3

Voltage = ?

Now,

R = V/A

11.7 = V/0.3

11.7*0.3 =V

V = 3.51

I hope it's help you...

Mark me as brainliest...

7 0

3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

$kA^2 =mV^2 +ky^2$

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

$1700*A^2=5.3*1.7^2 +1700*(0.045)^2$

Solving for A,

$A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}$

$A ^2 = 0.011035$

$A=0.105 m \approx 10.5 cm$

PART C) Finally, the total mechanical energy is given by the equation

$E = \frac{1}{2}kA^2$

$E=\frac{1}{2}1700*(0.105)^2$

$E= 9.3712 J$

3 0

3 years ago

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is

vodka [1.7K]

The sphere’s Electric potential energy is 1.6* $10^{-6}$ J

Given,

q=6. 5 µc, V=240 v,

We know that sphere’s Electric potential energy(E) = qV=6.5* $10^{-6} *240$ =1.6* $10^{-6}$ J

<h3>Electric potential energy</h3>

The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.

In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?

Learn more about Electric potential energy here:

brainly.com/question/24284560

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8 0

1 year ago