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uysha [10]
3 years ago
11

A potter's wheel moves from rest to an angu-

Physics
1 answer:
dusya [7]3 years ago
8 0

Answer:

angular acceleration =<u>final angular speed + initial angular speed</u>

                                                                     time

                                  =<u>0.6283rad/s +0</u>

                                            24.6s

                                   =0.6283/24.5

                                   =0.025rad/s^2

Explanation:

first angular speed is equal to angular velocity and we will convert the unit to rad/s so 1rev/sec=6.283rad/s then angular acceleration is the rate at which the angular velocity changes with the time we have the final angular velocity,initial angular velocity and time now substitute the numbers in the formula and the value will be 0.025 rad per seconde square

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

5 0
3 years ago
What is the period of a wave if 3 waves passes every 6 seconds
Naya [18.7K]
Time period = time/no. of waves = 6/3 = 2s
4 0
3 years ago
If 90 grams of gold has a volume of 5 cm3, calculate the density (include the units and show your work or you get no credit
ehidna [41]

Answer:

The density of gold is of 18 grams per cm3.

Explanation:

The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:

90/5 = X

18 = X

Thus, the density of gold is 18 grams per cm3.

6 0
3 years ago
g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin
DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as mgh=0.5mv^{2} since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then v=\sqrt {2gh} and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s

6 0
3 years ago
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest
yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}=v_{i}-gt (1)

Where:

  • v(i) is the initial velocity
  • v(f) is the final velocity
  • g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

0=30-9.81t

Solve it for t:

t=3.06\: s

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}^{2}=v_{i}^{2}-2gh (2)

0=v_{i}^{2}-2gh

We know that the initial velocity is 30 m/s.

0=30^{2}-2gh

Solving it for h we have:  

h=\frac{30^{2}}{2*9.81}

h=45.87 \: m

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

10=30-(9.81t)

So, the time elapsed to get 10 m/s is:

t_{upward}=2.04\: s

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

t_{upward}=2.04+t  

t=1.02\: s

This is the time when the ball has 10 m/s downward.          

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

t=2t_{upward}

t=2*3.06      

t=6.12\: s

The displacement will be zero after 6.12 s.  

e) Here we need to find the time when v(f) is 15 m/s

15=30-gt

t=\frac{15}{9.81}  

t=1.53\: s

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0
3 years ago
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