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2 years ago

8

Is the frictional force the same as the applied force when the net force equals zero?

1 answer:

DerKrebs [107]2 years ago

6 0

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

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A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the

Valentin [98]

To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
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x = 0.32 m

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3 years ago

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A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition

Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

$\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}$

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

$\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\$

<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>

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3 years ago

Can someone help me?!!!!!

ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

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Scalar: -6

so the vector multiplied by the scalar will have components

$(-3\cdot (-6), -2 \cdot (-6))=(18,12)$

The magnitude is given by Pythagorean's theorem:

$m=\sqrt{18^2+12^2}=21.6$

and the direction is given by the arctan of the ratio between the y-component and the x-component:

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3 years ago

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2 years ago

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A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the

adelina 88 [10]

Answer:

maximum possible velocity = $\sqrt{ugR}$

Explanation:

centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.

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where v is the velocity of car and r is the radius of circle

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where u is the coefficient of static friction.

therefore $umg\geq \frac{mv^{2}}{R}$

therefore maximum possible velocity = $\sqrt{ugR}$

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3 years ago