Who think's im pretty <br> the pic is a lil hard to see

daser333 [38]

The answer is C..........

4 0

3 years ago

A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at

docker41 [41]

Answer:

$y = 4.36$

Explanation:

Let the height of the ladder be L

$L = 10$

Also:

Let $x = distance\ from\ the\ base\ of\ the\ ladder$

Let $y = height\ of\ the\ base\ of\ the\ ladder$

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

$L^2 = x^2 + y^2$ --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

$x = 9$

Substitute 9 for x and 10 for L in $L^2 = x^2 + y^2$

$10^2 = 9^2 + y^2$

$100 = 81 + y^2$

Make $y^2$ the subject

$y^2 = 100 - 81$

$y^2 = 19$

Make y the subject

$y = \sqrt{19$

$y = 4.36$

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0

3 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

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

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

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

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

3 years ago

Read 2 more answers

How does the force of gravity between two bodies change when the distance between them doubles? 1. unable to determine; the mass

Rzqust [24]

6. Drop to one quarter of its original value

7 0

3 years ago

4:list one food chain that is part of the food web.

Sophie [7]

4. Grass - Caterpillar - Hedgehog - Fox
5. Caterpillar, Rabbit, Mouse.
6. Cougar and Fox.
7. Bacteria
8. The bird, hedgehog, Fox and cougar would be effected since the Hedgehogs and birds would soon die out due to the loss of their food. Once they die out, the cougar and Fox would have no predators left to eat.

6 0

3 years ago

Who think's im pretty the pic is a lil hard to see