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Nitella [24]
2 years ago
12

A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of

the stone if it hit the plane a distance L down the slope from the point of launch?
Physics
1 answer:
Katen [24]2 years ago
7 0

Answer:

S = V t     where S is the horizontal distance traveled

1/2 g t^2 = H       where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

Giving V in terms of L and theta

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force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it
Yanka [14]

Answer:

Given force=10lb

L1=4in converting to feet

But 0.08333ft= 1 inch

Then 4 inch is 0.3332

6inch is 0.49998

But hookes law states

F=Kx where F is force,K is the force constant ,X

K=F/X=10/0.3333=30N/m

Integrating this

Integral of 30x with limit 0.333 to 0.5

F=30x^2/2=15x^2substing the limit

F=(15(0.5^2-0.33^2)=2.08ft-lb

Explanation:

3 0
3 years ago
Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False
kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

\vec{L} = I \vec{\omega}

where I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2 is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2} so \vec{\omega} \neq 0 unless \vec{r} ║  \vec{v}.

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0
3 years ago
Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c
grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

The static and dynamic friction coefficient is, \mu=0.5

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

F_A-\mu mg=ma

\dfrac{F_A-\mu mg}{m}=a

a=\dfrac{1000-0.5(1000)(9.81)}{1000}

a=-3.905\ m/s^2

So, the acceleration of the car is -3.905\ m/s^2. Hence, this is the required solution.

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