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Vikentia [17]
3 years ago
7

An empty office chair is at rest on a floor. Consider the following forces:. 1. A downward force due to gravity;. 2. An upward forc

e exerted by the floor; and. 3. A net downward force exerted by the air. Which force(s) act on the office chair?. 1. None of the forces act on the chair since. it is at rest. 2. All three forces. 3. 1 only. 4. 1 and 2 only. 5. 2 and 3 only.
Physics
1 answer:
Ghella [55]3 years ago
7 0
The answer is:

<span> 4. 1 and 2 only.</span>
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A disoriented physics professor drives 3.22 km north, then 4.75 km west, then 1.90 km south. find the magnitude and direction of
Andre45 [30]

It can be noted that the directions west and north are perpendicular to each other. hence we can use the right angle formula for calculating total distance in these directions. Also South is just negative north. This technique is called euclidean distance norm. It is the basis of Cartesian coordinate system.

Total distance = 3.22 N + 4.75 W +1.90  S

= 1.32 N + 4.75 W

D = \sqrt{N^2 + W^2}

D = 4.93 Km

direction = atan(N/W) = 15.53 deg north of west

7 0
3 years ago
The __________-second rule applies to any speed in ideal weather and road conditions.
SOVA2 [1]
The two-second rule.

It is a common guideline to follow while driving.

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8 0
3 years ago
HELP PLEASE DUE IN 3 MINUTES
diamong [38]

Answer:

tectonic plate movement

Explanation:

6 0
2 years ago
Calculate the force that the 4kg block exerts on the 10kg block
Kryger [21]
Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.
8 0
3 years ago
You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the
podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

F_x = Fcos30


F_y = Fsin30


Now the normal force on the block is given as

N = Fsin30 + mg

N = 0.5F + (18\times 9.8)

N = 0.5F + 176.4

now the friction force on the cart is given as

F_f = \mu N

F_f = 0.625(0.5F + 176.4)

F_f = 110.25 + 0.3125F

now if cart moves with constant speed then net force on cart must be zero

so now we have

F_f + F_x = 0

Fcos30 - (110.25 + 0.3125F) = 0

0.866F - 0.3125F = 110.25

F = 199.2 N

so the force must be 199.2 N

8 0
3 years ago
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