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Gekata [30.6K]
3 years ago
6

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

egrees.
Physics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

v=18.89\frac{m}{s}

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)

Solving N from (2) and replacing in (1):

N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Replacing and solving for v:

gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}

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Answer:

Due to suffocation and toxic paint.

Explanation:

The population of hermit crabs has been decreasing because of suffocation and poison. Many crabs actually die slowly from suffocation because their gills require high humidity in order to breathe properly. Hermit crabs are slowly poisoned by tap water in which the toxic substances present in the paint that dissolves their shells that leads to the death of hermit crabs.

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3 years ago
A baseball player throws a baseball with a velocity of 13 m/s north. It is caught by a second player seven seconds later. How fa
guajiro [1.7K]

Answer:

A. 91 meters north

Explanation:

Take +y to be north.

Given:

v₀ = 13 m/s

a = 0 m/s²

t = 7 s

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Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²

Δy = 91 m

The displacement is 91 m north.

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3 years ago
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Answer:

Explanation:

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3 years ago
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

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3 years ago
An object covers a distance of 8 meters in the first second of travel, another 8 meters
andre [41]

Answer:

I guess the acceleration would be 8 meters a second

Explanation:

I can't think of any other fitting way to put the answer sorry if it's not right

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3 years ago
Read 2 more answers
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