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2 years ago

A wooden block is let go from a height of 5.80 m. What is the velocity of the block just before it hits the ground?

Physics

1 answer:

Shalnov [3]2 years ago

3 0

Given :

A wooden block is let go from a height of 5.80 m.

To Find :

The velocity of the block just before it hits the ground.

Solution :

We know, by equation of motion :

$v^2 - u^2 = 2as$

Here, a = g = 9.8 m/s²( Acceleration due to gravity )

Putting all given values in above equation, we get :

$v^2 - u^2 = 2as\\\\v^2 -0 = 2\times 9.8 \times 5.8 \\\\v = \sqrt{2\times 9.8 \times 5.8 } \ m/s\\\\v = 10.66\ m/s$

Hence, this is the required solution.

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A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

$\omega=0.37 [rad/s]$

I hope it helps you!

5 0

3 years ago

What is true of electrolytic cells but is not true of galvanic cells?

ira [324]

They are non-spontaneous

8 0

3 years ago

Read 2 more answers

The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler

guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

$v^{2}=u^{2}+ 2as$

$620^{2}=0^{2}+ 2\times a \times 0.89$

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.

6 0

3 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)