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Zina [86]
3 years ago
8

Can someone list two non-examples of volcano formations/eruptions?

Physics
1 answer:
Svet_ta [14]3 years ago
6 0
Water and Soil i think
You might be interested in
PLEASE help me!!! ASAP!! Ill mark you as brainliest!!!!
MakcuM [25]

Answer:

B

Explanation:

graph b shows a steady pace of movement for 20 minutes and then shows a plateau in the distance, showing that while time keeps moving (obviously), the distance doesn't change. then after 5 minutes, the student gets up and starts running again. hope this helped!

8 0
3 years ago
A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.0 cm from the axis of rotation. (a) Calcul
Annette [7]

Answer:

a) a = 0.477 m/s^2

b) u = 0.04862

Explanation:

Given:-

- The rotational speed of the turntable N = 33 rev/min

- The watermelon seed is r = 4.0 cm away from axis of rotation.

Find:-

(a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip

Solution:-

- First determine the angular speed (w) of the turntable.

                   w = 2π*N / 60

                   w = 2π*33 / 60

                   w = 3.456 rad/s

- The watermelon seed undergoes a centripetal acceleration ( α ) defined by:

                   α = w^2 * r

                   α = 3.456^2 * 0.04

                   α = 0.477 m / s^2

- The minimum friction force (Ff) is proportional to the contact force of the seed.

- The weight (W) of the seed with mass m acts downwards. The contact force (N) can be determined from static condition of seed in vertical direction.

                   N - W = 0

                   N = W = m*g

- The friction force of the (Ff) is directed towards the center of axis of rotation, while the centripetal force acts in opposite direction. The frictional force Ff = u*N = u*m*g must be enough to match the centripetal force exerted by the turntable on the seed.

                    Ff = m*a

                    u*m*g = m*a

                    u = a / g

                    u = 0.477 / 9.81

                    u = 0.04862

8 0
3 years ago
A race car accelerates from 16.5 m/s to 45.1 m/s in 2.27 seconds. Determine the acceleration of the car.
ioda

Answer:

12.6

Explanation:

4 0
3 years ago
A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up
svlad2 [7]

Answer:

W = (F1 - mg sin θ) L,   W = -μ  mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis  

       N - W_{y} =

       N = W_{y}

X axis

       F1 - fr - Wₓ = 0

       fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     cos θ = W_{y} / W

      Wₓ = W sin θ

      W_{y} = W cos θ

We substitute

      fr = F1 - W sin θ

Work is defined by

        W = F .dx

        W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

         

        W = -fr x

        W = (F1 - mg sin θ) L

Another way to calculate is

         fr = μ N

         fr = μ W cos θ

the work is

         W = -μ  mg cos θ L

4 0
3 years ago
A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.
crimeas [40]

Answer:

Explanation:

Given:

length of ladder r_L = 14m

weight of ladder F_L = 490N

position of firefighter r_F = 3.8m

weight of firefighter F_F = 820N

angle of ladder \alpha = 63

Unknown:

force of the wall on the ladder F_W

force of friction on base of ladder F_R

normal force on base of ladder F_N

From the free body diagram of the sketch you get 3 equations:

F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0

Solving the equations gives:

F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}

a)

F_R = 238N\\ F_N = 1310N

b)

F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3

c) Using the result from b and solving for r_F

\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m

4 0
4 years ago
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