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asambeis [7]
3 years ago
15

A cart loaded with bricks has a total mass of 25.1 kg and is pulled at constant speed by a rope. The rope is inclined at 21.3 ◦

above the horizontal and the cart moves 31 m on a horizontal floor. The coefficient of kinetic sommer (irs359) – WEP Quiz – craig – (2019-1A AP12) 2 friction between ground and cart is 0.5 . The acceleration of gravity is 9.8 m/s 2 . What is the normal force exerted on the cart by the floor? Answer in units of N. 011 (part 2 of 3) 10.0 points How much work is done on the cart by the rope? Answer in units of kJ. 012 (part 3 of 3) 10.0 points Note: The energy change due to friction is a loss of energy. What is the energy change Wf due to friction?
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
5 0

Answer:

2.0 m Since we've been told to ignore friction, then we can also ignore gravity since all it's doing is contributing to the normal force which is only of interest if we're paying attention to friction. So how fast will 13 N of force accelerate 34 kg of mass? Since a newton is kg*m/s^2 and we want to get m/s^2, a simple division should be able to cancel out the kg. So: 13 kg*m/s^2 / 34 kg = 0.382352941 m/s^2 The distance an object moves under constant acceleration is d = 0.5 AT^2 So let's substitute the known values and calculate the distance. d = 0.5 AT^2 d = 0.5 0.382352941 m/s^2 * (3.2 s)^2 d = 0.191176471 m/s^2 * 10.24 s^2 d = 1.957647059 m Rounding to 2 significant figures gives 2.0 meters

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You throw a balloon that floats in the air with a velocity of 2 m / s south . If the wind speed is 5 m / s west , how far south
zvonat [6]

Answer:

The distance traveled by the balloon is 10.77 m

Explanation:

velocity of the ball, v_b = 2 m/s south

velocity of the air, v_a = 5 m/s west

To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.

                                       | 2m/s

                                       |

                                       |

                                      ↓

      5m/s  ←------------------

the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.

R² = 5² + 2²

R² = 29

R = √29

R = 5.385 m/s

The distance traveled by the balloon is calculated as;

d = R x t

where;

t is time of the motion = 2 seconds

d = 5.385 x 2

d = 10.77 m

Therefore, the distance traveled by the balloon is 10.77 m.

4 0
3 years ago
If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli
Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

There are 2*pi radians in 1 revolution so:

RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

So you need about 30 RPM to keep people from falling out the bottom

7 0
3 years ago
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6 0
3 years ago
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