The surface tension of isopropanol in air has a value of 23.00 units and the

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

(c) Magnitude of the force that keeps you go around at this acceleration

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

2 years ago

Vector A has magnitude of 15.0 m/s and is 75° counter-clockwise up from the x-axis. What are the x- and y-components of the vect

LiRa [457]

Answer:

x component 3.88 y- component 14.488

Explanation:

We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure

Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )

y- component will be 15 sin 75°=14.488

For verification the resultant of x and y component should be equal to 15

So $resultant =\sqrt{14.488^2+3.88^2}=15$

3 0

3 years ago

a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to

Natalka [10]

Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

4 0

3 years ago

An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30.0 s before leaving the ground. a. How far di

rosijanka [135]

The formula we can use in this case is:

d = v0t + 0.5 at^2

v = at + v0

where,

d = distance travelled

v0 = initial velocity = 0 since at rest

t = time travelled

a = acceleration

v = final velocity when it took off

a. d = 0 + 0.5 * 3 * 30^2

d = 1350 m

b. v = 3 * 30 + 0

8 0

3 years ago

Is the matching correct?

iragen [17]

Yes thats right :)

have a great day!!!

7 0

3 years ago