What is the function of a hypothesis in the scientific inquiry process

Explain why, if you go to the moon, your weight would change but your mass wouldnt

Evgen [1.6K]

There's little gravity so your weight would change but not your mass

3 0

3 years ago

Hurricanes are considered____ because they lose power over cool waters or land A. Short-lived B. Heat engines C. Weak

klasskru [66]

My best guess would be heat engines

7 0

3 years ago

Read 2 more answers

If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 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

Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)

Gala2k [10]

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

8 0

3 years ago