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3 years ago

Describe the forces involved in Throwing a spear

Physics

1 answer:

Alexeev081 [22]3 years ago

8 0

Explanation:

The center of gravity is near the grip and does not change during throw. "Throwing through the tip," a popular term of how to throw a javelin, means throwing through the grip or center of gravity. The center of pressure is the aerodynamic force of drag and lift on the javelin.

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Force is applied to an object and the object is moved over a distance in the same direction of the applied force" is the definit

just olya [345]

Answer: The correct answer is "work".

Explanation:

The expression for the work done in terms of force and displacement is as follows;

W= Fd

Here, W is the work done, F is the applied force and d is the displacement.

If the force is applied to an object and the object is displaced to some distance then the work is done.

It is given in the problem that force is applied to an object and the object is moved over a distance in the same direction of the applied force". The given statement is the definition of work. The work done can be zero, negative and positive. In the given case, the positive work is said to be done as the direction of the applied force and the distance moved by the object are same.

Therefore, the correct option is (d).

7 0

3 years ago

Read 2 more answers

When ionic compounds are made using transition metals, ______________ are used in their naming.

forsale [732]

Answer:

Romen numerals

Explanation:

6 0

4 years ago

The si unit of mass is the Newton. True or false.

LUCKY_DIMON [66]

False it’s the unit of force

5 0

3 years ago

The lever PQ is welded to the bent rod QSTU which is supported by a single thrust bearing at S and by the rigid link UV. Link UV

valentinak56 [21]

Answer:

The answer to the questions are as follows

a) The magnitude of the force in UV = 512.41 N

b) The five support reaction components are

1. PQ turning moment force about the z axis

2. PQ turning moment force about the y axis

3. UV turning moment force about the y axis

4. UV turning moment force about the z axis

5. UV turning moment force about the x axis

Please see below

Explanation:

When the system is in 3D equilibrium

Sum of forces in x direction = 0

Sum of forces in y direction = 0

Sum of forces in z direction = 0

Also, Sum of moments in x direction = 0

Sum of moments in y direction = 0

Sum of moments in z direction = 0

From the diagram, we have force in the +ve x direction = 600 N

Also moment about y = 600×0.08 = 48 Nm

Moment about z = 600×0.09 = -54 Nm

Moment about x, = 0 Nm

For the support S, taking moment about x = 0.05 × S

taking moment about y = 0

taking moment about z = 0

At point U, taking moment about y = -0.1U

moment about x = -0.05U

moment about z = -0.1U

Summing the moments we have

In the x, y and z directions

0 + 0.04Vy-1.46Vz -0.05Uz= 0.... (1)

48 + 0.1Vz-0.04Vx - 0.1Uz = 0... (2)

-24 + 0.146Vx-0.1Vy + 0.05Ux + 0.01Uy = 0.... (3)

Ux + Vx = 600 ....(4)

Uy = -Vy .....(5)

Uz = -Vz .......(6)

Solving the above 6 equations we get the following values

Ux = 242.28 N

Uy = -366.74

Uz = -10.40

Vx = 357.72N

Vy = 366.74N

Vz = 10.40N

The magnitude of the force in UV = $\sqrt{357.72^{2} + 366.74^{2} + 10.40^{2} }$ = 512.41 N

b) The five support reaction components are

1. PQ turning moment force about the z axis

2. PQ turning moment force about the y axis

3. UV turning moment force about the y axis

4. UV turning moment force about the z axis

5. UV turning moment force about the x axis

5 0

4 years ago

4. The length of a simple pendulum whose time period is 2 second ?<br>

Scilla [17]

Answer:

We know that the time period of the vibration of the second's pendulum is T=2s. Hence, the length of the second's pendulum is 99.3 cm

3 0

2 years ago