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andre [41]
2 years ago
5

A given wave has a wavelength of 1.4 m and a frequency of 2.0 Hz. How fast

Physics
1 answer:
stiks02 [169]2 years ago
3 0

Answer:

The wave speed is 2.8 m.

Explanation:

Wavelength = 1.4 m

frequency, f = 2 Hz

the wave speed is given by

wave speed = wavelength x frequency

wave speed = 1.4 x 2 = 2.8 m

option (D) is correct.  

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The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th
sertanlavr [38]

Answer:

The force exerted by the biceps is 143.8 kgf.

Explanation:

To calculate the force exerted by the biceps, we calculate the momentum in the elbow.

This momentum has to be zero so that her forearm remains motionless.

Being:

W: mass weight (6.15 kg)

d_W= distance to the mass weight (0.425 m)

A: weight of the forearm (2.25 kg)

d_A: distance to the center of mass of the forearm (0.425/2=0.2125 m)

H: force exerted by the biceps

d_H: distance to the point of connection of the biceps (0.0215 m)

The momemtum is:

H*d_H-A*d_A-W*d_W=0\\\\H=(A*d_A+W*d_W)/d_H\\\\H=(2.25*0.2125+6.15*0.425)/0.0215\\\\H=(0.478125+2.61375)/0.0215\\\\H=3.091875/0.0215=143.8

The force exerted by the biceps is 143.8 kgf.

8 0
3 years ago
What is the purpose of a circuit breaker?
sashaice [31]
It’s designed to protect an electrical circuit from damage caused by overcurrent, usually resulting from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.

That’s really just the basic purpose.
Happy to help!
~Brooke❤️
8 0
3 years ago
A 1500 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2500 kg SUV traveling from ea
kolezko [41]

Answer:

v = 19.33 m / s   South

Explanation:

To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.

Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction

initial instant. Before the crash

        p₀ = m 0 + M v₂ₓ

final instant. Right after the crash

        p_f = (m + M) vₓ

        p₀ = 0_pf

        M v₂ₓ = (m + M) vₓ

In this case m is the mass of the car and M the mass of the SUV

        vₓ = \frac{M}{m+My}  v₂ₓ            (1)

in the Y axis (North - South direction)

initial instant

       p₀ = m v_{1y} + M 0

final moment

       p_f = (m + M) v_y

       p₀ = p_f

       m v_{1y} + M 0 = (m + M) v_y

       v_y = \frac{m}{m+M}  \  v_{1y}       (2)

With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.

         W = ΔK

friction force work is

         W = - fr d

the friction force is described by the equation

         fr = μ N

Newton's second law

         N-W = 0

         N = W

         

we substitute

         W = - μ (m + M) g d

as the car stops the final kinetic energy is zero and

the initial kinetic energy is

         K₀ = ½ (m + M) v²

we substitute

         - μ (m + M) g d = 0 - ½ (m + M) v²

            μ g d = ½ v²

            v² = 2 μ g d

the distance traveled can be found with the Pythagorean theorem

        d = \sqrt{x^2+y^2}

        d = \sqrt{5.48^2 + 6.37^2}

        d = 8.40 m

let's calculate the speed

         v² = 2 0.75 9.8 8.40

         v = √123.48

         v = 11.11 m / s

this velocity is in the direction of motion so we can use trigonometry to find the angles

          tan θ = y / x

          θ = tan⁻¹ y / x

          θ = tan⁻¹ (-5.48 / -6.37)

          θ = 40.7º

Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.

          θ'= 180 + 40.7

          θ’= 220.7º

In the exercise they indicate the the sedan moves in the y-axis, therefore

          sin θ'= v_y / v

          v_y = v sin 220.7

          v_y = 11.11 sin 220.7

          v_y = -7.25 m / s

the negative sign indicates that it is moving south

To find the speed we substitute in equation 2

          v_y = \frac{m}{m+M}  \ v_{1y}

          v_{1y} = v_ y   \frac{m+M}{m}

           

let's calculate

         v_{1y} = -7.25    \frac{1500+2500}{1500}

         v_{1y} = - 19.33 m/s

therefore the speed of the sedan is v = 19.33 m / s with a direction towards the South

4 0
2 years ago
Which is correct about the charge movement when a battery is connected to copper wire?a.)Electrons leave the positive battery te
professor190 [17]

Answer:

b.)Electrons leave the negative battery terminal and electrons enter the positive terminal.

Explanation:

- First of all, the carriers of charge in an electrical circuit are the electrons, not the protons. In fact, electrons in a conductor (such as the copper wire) are free to move, while protons are bond inside the nuclei of the atoms, so they are not free to move. This means that only choices a) and b) could be correct.

- Finally, we know that charges with same sign repel each other, while charges with opposite sign attract each other. This means that electrons are repelled by the negative terminal of the battery and attracted by the positive terminal of the battery. Therefore, the correct choice must be

b.)Electrons leave the negative battery terminal and electrons enter the positive terminal.

5 0
3 years ago
What is the effective resistance between the points A and D? A) 1.3 Ω B) 2.2 Ω C) 10 Ω D) 12 Ω
Rudiy27

Answer:

B) 2.2 Ω

Explanation:

First of all, we should notice that the resistor placed between A and B is short-circuited. In fact, the current from point A will follow the wire above between A and C (which has zero resistance), so we can basically ignore the presence of the resistor between A and B, since it has no effect on the circuit.

Then, we can notice that the two resistors between CB and CD are in parallel to each other; therefore, their equivalent resistance is given by:

\frac{1}{R}=\frac{1}{R_{BC}}+\frac{1}{R_{CD}}=\frac{1}{4.0 \Omega}+\frac{1}{5.0 \Omega}=\frac{9}{20 \Omega}\\R= \frac{20 \Omega}{9}=2.2 \Omega

6 0
3 years ago
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