I need help with high school chemistry!!! Can someone help

A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating system = 6.74 J

We know that energy is given by

(a) $Ke=\frac{1}{2}mv_{max}^2$

$6.74=\frac{1}{2}\times 0.267\times v_{max}^2$

$v_{max}=7.1m/sec$

(b) Now $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec$

We know that $\omega =\sqrt{\frac{k}{m}}$

$35.68=\sqrt{\frac{k}{0.267}}$

$k=339.9N/m$

(c) We know that energy is given by

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

$6.74=\frac{1}{2}\times 339.9\times A^2$

$A=19.91cm$

4 0

2 years ago

Assuming no air resistance, how far will a 0.0010 kg raindrop have fallen when it hits the ground 30.0 s later. Assume g = 9.8 m

Serga [27]

D = (1/2)·at²
where d is the distance fallen, a is the acceleration (g in this problem), and t is the time
d = (1/2)·(9.8 m/s²)·(30 s)² = (1/2)·(9.8)·(900) m
d = 4410 m
The answer is b) 4410 m

Note: the mass of the raindrop is irrelevant since the acceleration due to gravity is independent of mass. (Galileo's Leaning Tower of Pisa experiment)

6 0

3 years ago

Read 2 more answers

In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau

mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

Em₀ = U = m g y₁

final point. At the exit of the ramp

Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

Em₀ = Em_f

m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

v = w r

the moment of inertia of a sphere is

I = $\frac{2}{5}$ m r²

we substitute

m g (y₁ - y₂) = ½ m v² + ½ ( $\frac{2}{5}$ m r²) ( $\frac{v}{r}$ )²

m g h = ½ m v² (1 + $\frac{2}{5}$ )

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

mg h = $\frac{7}{5}$ ( $\frac{1}{2}$ m v²)

v = √5/7 √2gh

This is the exit velocity of the vertical movement of the sphere

v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

starting point

Em₀ = U = mg y₁

final point

Em_f = K + U = ½ m v² + m g y₂

Em₀ = Em_f

mg y₁ = ½ m v² + m g y₂

m g (y₁ -y₂) = ½ m v²

v = √2gh

this is the speed of the box

v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

v² = v₀² - 2 g y

at the highest point v = 0

y = vo₀²/ 2g

for the sphere

y_sphere = 5/7 2gh / 2g

y_esfera = 5/7 h

for the block

y_block = 2gh / 2g

y_block = h

therefore the block reaches higher than the sphere

$\frac{y_{sphere}} {y_bolck} = 5/7$

3 0

3 years ago

A compass taken to Earth's moon does not point in a specific direction on the moon.

yKpoI14uk [10]

C) the moon does not have a strong magnetic field

5 0

2 years ago

Read 2 more answers

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

3 0

3 years ago