Ask question

Ask question

All categories

kondor19780726 [428]

2 years ago

14

Particles 1 and 2 each mass m fixed to the ends of a rigid massless rod of length L1 + l2 with L1 = 20cm and l2 = 80 cm. The rod

is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of particles 1 and 2.

1 answer:

Degger [83]2 years ago

7 0

Answer:

Sorry bro I don't even know the answer

You might be interested in

3. A bottle of vitamin C contains 100 tablets and weighs 80 g. If the

Elan Coil [88]

Answer:

a

0.57g

b

20.52

c

28.5

Explanation:

a

The bottle weighs 80g with tablets

If the bottle alone weighs 23g, the tablets weigh 57g

100 tablets weigh 57g, 1 tablet weighs 0.57g

b

0.57g is one tablet, so to achieve 36 tablets we must multiply by 36

0.57 multiplied by 36 is 20.52

c

To find 50 tablets we can use the same method we used before or a slightly faster method

100 tablets is 57g, so all we have to do is halve to find 50

57 divided by 2 is 28.5

4 0

2 years ago

You have two balls of equal size and smoothness, and you can ignore air resistance. One is heavy, the other is much lighter. You

motikmotik

Answer:

They will hit the ground at the same time.

Explanation:

By ignoring the opposing forces i.e. air resistant, both the heavy and light balls will fall with same acceleration due to gravity (g=9.8 m/s²) and g is independent of mass of the objects. Thus both will hit the ground at the same time.

4 0

2 years ago

Which equation below represents a generic equation suggested by a graph showing a hyperbola?

ruslelena [56]

Answer:

y = k/x

Explanation:

y = k/x is a graph of a hyperbola that has been rotated about the origin.

8 0

2 years ago

A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca

Lana71 [14]

Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

$v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t$

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

5 0

3 years ago

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

charle [14.2K]

Answer:

She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

$77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

So she will cover 77 m in 2.85 seconds

Now considering vertical motion, up direction as positive

Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 $m/s^2$ , time = 2.85

Substituting

$s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

So she will make the jump.

6 0

2 years ago