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3 years ago

A trapeze artist swings in simple harmonic motion with a period of 3.8 s.

1 answer:

7 0

As we know that time period of simple pendulum is given as

T = 2π √L/g

here we know that

T = 3.8 s

now from above equation we know that

T² = 4π² (L/g)

now on rearranging the above equation we will have

L = gT² / 4π²

now plug in all data into it

L = (9.8) (3.8)² / (4) (3.14)²

so the length of the cable must be 3.6 m

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A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini

finlep [7]

<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁, object 2 be u₂

Final velocity object 1 be v₁, object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

Initial momentum = Final momentum

24 = 3v₁ + 6M

v₁ + 2M = 8

v₁ = 8 - 2M

Initial kinetic energy = Final kinetic energy

96 = 1.5 v₁² + 18 M

v₁² + 12 M = 64

Substituting v₁ = 8 - 2M

(8 - 2M)² + 12 M = 64

64 - 32M + 4M² + 12 M = 64

4M² = 20 M

M = 5 kg

Option 2 is the correct answer.

6 0

2 years ago

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago

While jumping on a trampoline you calculate that at the highest peak of your jump you have 900 joules of gravitational potential

BabaBlast [244]

Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.

<u>Explanation</u>

When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.

Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

8 0

3 years ago

Under which condition does Ohm's law apply? a. The current must be constant b. The power must be constant c. The temperature mus

ruslelena [56]

Answer:

option c

Explanation:

The temperature must be constant. Ohms law states that the current running through a conductor is directly proportional to the potential difference across it provided the temperature remains constant

6 0

2 years ago

A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz

Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, $\bar f$ is given by

$\bar f=f\frac {V}{V-V_s}$ where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and $V_s$ is the speed of car which is unknown, \bar f is given as 440 Hz

$440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s$

8 0

2 years ago