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mezya [45]
3 years ago
14

Detlev walks 1000 meters north to the store and walks back 900 meters south to his friends house. The friends house is 100 meter

s to the north of Detlev’s house. What was Detlev’s Travel Distance?
Physics
1 answer:
lilavasa [31]3 years ago
3 0
1000 is the correct answer
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Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch i
Olegator [25]

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

4 0
3 years ago
A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 mm behind the nose of the airplane, the front w
padilas [110]

Answer:

The percentage of the weight supported by the front wheel is  A= 19.82 %

Explanation:

B] Let the mass of plane be m, force on backwheels be Nb and nosewheel be Nn

Torque equation about nose wheel,

mg*(2.58-0.8) - Nb *(3.02-0.8) = 0

Nb =  mg*(2.58-0.8)/(3.02-0.8) = 0.8018 mg

Nn = mg - Nb = (1-0.8018) mg = 0.1982mg

Weight percentage supported by front wheel = 19.82% answer

6 0
2 years ago
A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a
Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

\Delta x=v_0t+\frac{gt^2}{2}

In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0
3 years ago
The direction of the acceleration of an object on a(n) _______________________ path is toward the _______________________ of the
aleksandrvk [35]

Answer:

circular...center

Explanation:

Physics

8 0
3 years ago
A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in
Novay_Z [31]

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

  • μ(s) is coefficient of static friction
  • m is mass of the block
  • g is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

#SPJ1

4 0
2 years ago
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