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3 years ago

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A pendulum is swinging back and forth with a period of 2.0 seconds here on Earth. This pendulum is then brought to the Moon, whe

re the acceleration due to gravity is much smaller. What will happen to the period of the pendulum, assuming everything else about it (mass, length, initial swing height, etc) remains exactly the same? Explain your answer.

1 answer:

Kryger [21]3 years ago

7 0

Answer:

A greater period.

Explanation:

The period of a simple pendulum is given by the following formula:

$T = 2\pi \cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Length of the pendulum

$g$ - Gravity constant

Since acceleration due to gravity is smaller in the Moon, period will be greater.

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How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai

Naddik [55]

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, $q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C$

Charge 2, $q_2=2.4\ \mu C=2.4\times 10^{-6}\ C$

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

$U=k\dfrac{q_1q_2}{r}$

$r=k\dfrac{q_1q_2}{U}$

$r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}$

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

8 0

3 years ago

Superman is flying 54.5 m/s when he sees

Nady [450]

348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations $d=v_{0} t+\frac{at^{2} }{2}$ and $v_{f} =v_{0} +at$ where $d$ is distance, $v_{0}$ is the initial velocity, $v_{f}$ is the final velocity, $t$ is time, and $a$ is aceleration.

Superman's initial velocity is $v_{0}=54.5\frac{m}{s}$ , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know $d$ , $v_{0}$ and $t$ , we have to find the aceleration $a$ in order to find $v_{f}$ .

From the equation $d=v_{0} t+\frac{at^{2} }{2}$ we have to clear $a$ , getting the equation as follows: $a=\frac{2(d-v_{0}t) }{t^{2} }$ .

Substituting the values:

$a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}$

To find $v_{f}$ we use the equation $v_{f} =v_{0} +at$ .

Substituting the values:

$v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}$

5 0

3 years ago

An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

So

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

So

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

8 0

3 years ago

What happens in the process of gravitational condensation?

Lerok [7]

Answer:

An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.

3 0

3 years ago

Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200

andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

ΔK.E = ΔP.E

K_2 - K_1 = P_1- P_2

Where, K_2: Kinetic energy of hammer head as it hits the I-beam

K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

P_1: Initial gravitational potential energy of hammer head

- The expression simplifies to:

K_2 = P_1

Where, 0.5*m*v2^2 = m*g*s12

v2 = √(2*g*s12) = √(2*9.81*3)

v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

Wnet = ( P_3 - P_1 ) + W_friction

Wnet = m*g*s13 + F*s23

n*s23 = m*g*s13 + F*s23

Where, n: average force the hammerhead exerts on the I-beam.

s13 = s12 + s23

Hence,

n = m*g*( s12/s23 + 1) + F

n = 200*9.81*(3/0.074 + 1) + 60

n = 81562 N

6 0

3 years ago