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IceJOKER [234]
4 years ago
10

A rock dropped on the moon will increase it's speed from 0 m/s to 8.15 m/s in about 5 seconds what is the acceleration of the ro

ck
Physics
1 answer:
Lunna [17]4 years ago
5 0

Using the formula:


a = (Vf - Vi) / t


Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:


a = (8.15 - 0.0) / 5

a = 1.63 m/s^2


If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

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If an object is in equilibrium, which of the following statements is NOT true?
Naddik [55]
Equilibrium is when the opposing forces or influences are balanced.  So it would be E!

5 0
4 years ago
If the wavelength of a wave of light is 4.57x10^-9 m what is the frequency
alexandr402 [8]

Answer:

6.5646\times 10^{16}~ \text{Hz}

Explanation:

\text{Frequency,}~ f= \dfrac{c}{\lambda} = \dfrac{3\times 10^8 }{4.57 \times 10^{-9}} = 6.5646\times 10^{16}~ \text{Hz}

5 0
3 years ago
True or false: The maximum tensile force a solid, cylindrical wire can withstand increases as the thickness of the wire increase
MaRussiya [10]

Answer:

True

Explanation:

The tensile stress, σ, on a solid cylindrical wire is given by the following relationship;

\sigma = \dfrac{F_t}{A_o}

Where;

F_t = The tensile force

A_o = The original cross sectional area of the cylindrical wire = π·R²

R = The radius of the wire

Therefore;

F_t = σ × A_o = σ × π × R²

Therefore, the tensile force is directly proportional to the square of the radius of the cylindrical wire, and as the radius of the wire increases, which is by increasing the thickness of the wire, the tensile force is largely increased

The correct option is; True.

7 0
3 years ago
Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, c
Rus_ich [418]

Answer:

Explanation:

(a) When the plate starts to spin:

Its angular velocity increases, so the angular acceleration is non zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

(b) When the plate rotates at constant angular velocity:

Its angular velocity is constant so the angular acceleration is zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

(c) When the plate sows to halt:

Its angular velocity decreases, so the angular acceleration is non zero( but negative). As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

4 0
3 years ago
A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the
LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

\rho_w = \frac{m_w}{V_w}

If we solve for mw we got m_w = \rho_w V_w  (2)

Replacing equation (2) into equation (1) we got:

B = \rho_w V_w g (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0
4 years ago
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