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Illusion [34]
3 years ago
7

Suppose a car of mass m is moving at a constant speed v of

Physics
1 answer:
SIZIF [17.4K]3 years ago
4 0

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}

Finally the angle is

\arcsin(v^2/(gR))

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Answer:increased

Explanation:

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This causes the apparent to be increased if measured using weighing machine.

considering upward direction to be positive

N-mg=ma

N=m(g+a)

where N=Normal reaction=Apparent weight

a=acceleration of Elevator

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3 years ago
What is the weight of a 225kg space probe on the moon and the acceleration of gravity on the moon is 1.62
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3 years ago
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kolbaska11 [484]

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4 years ago
The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports
Katarina [22]
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.

c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02 
 
Alternative solution:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Therefore, they have to fly 82.02 mph
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Determine the radial acceleration of the ultracentrifuge using calculations
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When something moves on a round track, the guidance of the something's velocity must continually switch. A switching velocity means that there must be an acceleration. This acceleration is horizontal to the guidance of the velocity. This is said as “the radial acceleration”, or “centripetal acceleration” ("centripetal" means "center searching"). The “radial acceleration” is equal to “the square of the velocity”, divided by “the radius of the circular path of the object”. The unit of the “centripetal acceleration” is m/s².

\text { Centripetal acceleration }=\frac{\text {velocity}^{2}}{\text {radius of motion}}

\mathrm{a}_{\mathrm{rad}}=\frac{V^{2}}{r}

where,

\text { and }=\text { radial, or centripetal, acceleration }(\mathrm{m} / \mathrm{s} ^2)

"v" = "velocity" (m/s) and "r" = "radius of motion of the object" (m)

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