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3 years ago

4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable

Physics

1 answer:

icang [17]3 years ago

8 0

If an icy surface means no friction, then Newton's second law tells us the net forces on either block are

• m = 1 kg:

∑ F (parallel) = mg sin(45°) - T = ma … … … [1]

∑ F (perpendicular) = n - mg cos(45°) = 0

Notice that we're taking down-the-slope to be positive direction parallel to the surface.

• m = 0.4 kg:

∑ F (vertical) = T - mg = ma … … … [2]

Adding equations [1] and [2] eliminates T, so that

((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a

(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a

==> a ≈ 2.15 m/s²

The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.

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A rubber toy duck is at rest on an inclined plane. When the angle of inclination of the plane is increased to 36.0°, the toy duc

ch4aika [34]

Answer

toy begins to slide when the plane is at an angle of 36.0°

at this point the toy has just overcome the static friction

$f_s = \mu_s N$

$f_s = \mu_s mgcos \theta$

since toy begins to slide

$f_s = mgsin \theta$

now on comparing

$mgsin \theta = \mu_s mgcos \theta$

$\mu_s = tan \theta$

$\mu_s = tan (36^0)$

$\mu_s = 0.726$

as the body is moving at constant velocity acceleration will be zero

so now,

$m g sin \theta - f_k = ma$

$m g sin \theta - \mu_k mg cos\theta = 0$

$\mu_k = tan \theta$

$\mu_k = tan (29^0)$

$\mu_k = 0.554$

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3 years ago

Once the crate in sample problem 4C is in motion, a horizontal force of 53 N keeps the crate moving with a constant velocity. Fi

liraira [26]

Answer: k = 5.4kg/m

where m is the mass of the crate.

Explanation:

The force of kinetic friction is described by the equation:

Ff = k*N

in the opposite direction in wich the object moves, where k is the coefficient of kinetic friction and N is the normal force, that is equal to the weight of the crate. N = m*g where m is the mass of the crate and g is the gravitational acceleration.

If the crate keeps moving with constant velocity, this means that the crate is not accelerating so there is no net force applied on the crate. Then the friction force should be equal in magnitude to the horizontal force of 53N (but with different sign)

then we have:

k*m*g = 53N

k*m = 53N/9.8m/s^2 = 5.4kg

k = 5.4kg/m

In the question we do not have the mass of the crate, so you must put the value in that equation to get the value of k.

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3 years ago

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mariarad [96]

1.73 divide by 1.20=1.4416

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Kayla drew two ray diagrams to compare the behavior of light rays.

Naddika [18.5K]

The answer is the last option

Explanation:

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Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is

Leno4ka [110]

Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.

This led to Kepler's law of orbital motion.

First Law: Planetary orbits are elliptical with the sun at a focus.

Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.

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It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.

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4 years ago