Which best explains the relationship between heat energy and temperature?

A pendulum takes 10 seconds to swing through 2 complete cycles what is it’s period and frequency

Orlov [11]

its period should be the amount it takes to cycle from cycle to cycle so it would be 10 and your frequency would have to be calculated by taking 10 and dividing by 2 since that is how many cycles you have done so your frequency is 5

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7 0

3 years ago

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Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

3 years ago

Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons

dybincka [34]

The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = 229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N

3 0

2 years ago

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A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in t

arsen [322]

Answer:

v = 10.47 m/s

a = 219.32 m/s²

Explanation:

The velocity of the stone can be given by the following formula:

v = r ω

where,

v = linear velocity of the stone = ?

r = radius of tire = 0.5 m

ω = angular velocity = (200 rev/min)(2π rad/1 rev)(1 min/60 s) = 20.94 rad/s

Therefore,

v = (0.5 m)(20.94 rad/s)

v = 10.47 m/s

The acceleration of the stone will be equal to the centripetal acceleration. Therefore,

a = v²/r

a = (10.47 m/s)²/0.5 m

a = 219.32 m/s²

8 0

3 years ago

Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat

Natasha_Volkova [10]

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

The motorized toy boat moves at 8 m/s (u) at 4 s and at 8 m/s (v) at 8 s. We can calculate the acceleration (a) in that period using the following kinematic expression.

$a = \frac{v-u}{\Delta t} = \frac{8m/s-8m/s}{8s-4s} = 0 m/s^{2}$

The object with a mass (m) of 2.0 kg experiences an acceleration of 0 m/s². We can calculate the net force (F) in that period using Newton's second law of motion.

$F = m \times a = 2.0 kg \times 0 m/s^{2} = 0 N$

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

Learn more: brainly.com/question/13447525

6 0

2 years ago