652-05-01-01-00_files/i0190000.jpg What does the direction of arrow B indicate?

A 7.0-μC point charge and a point charge are initially extremely far apart. How much work does it take to bring the point charge

vazorg [7]

Answer:1.008 ×10^-14/rJ

Where r is the distance from.which the charge was moved through.

Explanation:

From coloumbs law

Work done =KQq/r

Where K=9×10^9

Q=7×10^-6C

q=e=1.6×10^-19C

Micro is 10^-6

W=9×10^9×7×10^-6×1.6×10^-19/r=100.8×10^-16/r=1.008×10^-14/rJ

r represent the distance through which the force was used to moved the charge through.

4 0

2 years ago

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An elevator is moving is an upwards

Minchanka [31]

Answer:

The elevator's free-body diagram has three forces, the force of gravity, a downward normal force from you, and an upward force from the tension in the cable holding the elevator. The combined system of you + elevator has two forces, a combined force of gravity and the tension in the cable.

Explanation:

6 0

3 years ago

. At what velocity will the box in Problem 6 be traveling when it hits the ground? Use formula: Va = Vf - Vi / 2

VladimirAG [237]

Answer:

How much time does his victim on the ground below have to move out of harm's way? At what velocity will the safe hit the ground? sownt d-200m. n a = 100/2. Vi-o.

Explanation:

5 0

2 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:

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

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3 years ago

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A mass of M-kg rests on a frictionless ramp inclined at 30°. A string with a linear mass density of μ=0.025" kg/m" is attached t

I am Lyosha [343]

Answer:

44.3 m/s

Explanation:

a) Draw a free body diagram of the mass M. There are three forces:

Weight force mg pulling down,

Normal force N pushing perpendicular to the ramp,

and tension force T pulling parallel up the ramp.

Sum of forces in the parallel direction:

∑F = ma

T − Mg sin 30° = 0

T = Mg sin 30°

T = Mg / 2

Draw a free body diagram of the hanging mass m. There are two forces:

Weight force mg pulling down,

and tension force T pulling up.

Sum of forces in the vertical direction:

∑F = ma

T − mg = 0

T = mg

Substitute:

mg = Mg / 2

m = M / 2

M = 2m

b) Velocity of a standing wave in a string is:

v = √(T / μ)

T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N. Therefore:

v = √(49 N / 0.025 kg/m)

v = 44.3 m/s

7 0

3 years ago