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Ann [662]
3 years ago
10

A dogsled team is shown pulling a man on a sled. Below the picture is a free body diagram with 4 force vectors. The first vector

is pointing downward, labeled F Subscript g Baseline = negative 50 N. The second vector is pointing left, labeled F Subscript f Baseline = negative 225 N. The third vector is pointing upward, labeled F Subscript N Baseline = 50 N. The fourth vector is pointing right, labeled F Subscript t Baseline = 225 N. The up and down vectors are the same length. The right and left vectors are the same length. Using the free-body diagram, calculate the net force acting on the sled. Is the sled in a state of dynamic equilibrium?
Physics
2 answers:
Scorpion4ik [409]3 years ago
8 0

Answer:

yes

Explanation:

its on edg.

Bess [88]3 years ago
7 0

Answer:

yes the sled is in a state of dynamic equilibrium.

Explanation:

the answer is yes because both sets of the vectors are the same length and the definition of <u>dynamic equilibrium</u> is: <u>the state in which an object in motion has a net force of zero</u>

<em><u>also if your using this to cheat DO IT YOURSELF.</u></em>

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Explanation:

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Rus_ich [418]

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7 0
3 years ago
Which physical property causes you to learn to one side when the bus you are traveling in takes a sharp turn?
Licemer1 [7]

Answer:

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3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
Someone help me! you get 27 points if you get it right
pashok25 [27]

Answer:

1. D

2. D

3. A

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3 years ago
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