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Jlenok [28]
1 year ago
8

Settlers are trying to determine the value of g on a distant planet. They throw a wrench downward from a height of 3 m. If the i

nitial speed of the wrench is 2 m/s, and the speed right before impact is 10 m/s, what is the value of g?
Physics
1 answer:
Agata [3.3K]1 year ago
8 0
<h2>Value of g on the distant planet is 16 m/s²</h2>

Explanation:

We have equation of motion v² = u² + 2as

Here initial velocity, u = 2 m/s

Final velocity, v = 10 m/s

Displacement, s = 3 m

We need to find acceleration, a.

Substituting

               v² = u² + 2as

               10² = 2² + 2 x a x 3

                6a =  96

                  a = 16 m/s²

Value of g on the distant planet is 16 m/s²

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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea
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Answer:

v = 3369.2 m/s

Explanation:

As we know that Beacon is rotating with angular speed

f = 10 rev/min

so we have

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\omega = 1.047 rad/s

now we know that

v = r \omega

here we will have

r = 2 miles = 2(1609 m)

r = 3218 m

so we have

v = 3218(1.047)

v = 3369.2 m/s

6 0
2 years ago
How long does it take for the velocity of the rain drop to reach 99% of its terminal velocity? (assume the conditions from part
vodomira [7]
If you think about it its part a and b 
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Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,
VARVARA [1.3K]

Answer: yes.

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It depends on 3 factors:

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3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer. 

6 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

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8 0
2 years ago
A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa
gavmur [86]

Answer : The correct option is, (d) 535^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of copper = 0.10cal/g^oC

c_2 = specific heat of water = 1.00cal/g^oC

m_1 = mass of copper = 120 g

m_2 = mass of water = 300 g

T_f = final temperature of mixture = 35^oC

T_1 = initial temperature of copper = ?

T_2 = initial temperature of water = 15^oC  

Now put all the given values in the above formula, we get:

120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC

T_1=535^oC

Therefore, the temperature of the kiln was, 535^oC

7 0
2 years ago
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