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DIA [1.3K]
4 years ago
12

If the distance between the two objects is reduced in half, what will be the changed force of attraction between them?

Physics
1 answer:
NNADVOKAT [17]4 years ago
8 0
Will reduce 4 times.

F =  G \frac{m1* m2}{ d^{2} }
So called LUG (Law of Universal gravitation).
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Mangrove forests play an important role in _______.
AysviL [449]
D. All of the above

At high tide fish will feed among the mangrove roots - rich fishing ground

The trees trap sediment and soil in the river that would flow out to sea which also helps stop erosion

Wildlife utilise almost every part of the tree, with insects and birds, monkeys and lizards in the branches, shrimps and fish in the roots, and snails and clams in the soil
4 0
2 years ago
In a ……………………………….. wave, the particles of the medium vibrate in a direction parallel
lapo4ka [179]

Answer:

4 longitudinal waves

Explanation:

have a beautiful day ahead

3 0
3 years ago
All of the following processes are the ones that lead directly to the formation of sedimentary rock EXCEPT-
emmasim [6.3K]

Answer:

melting of rock deep underground.

Explanation:

The melting of rocks deep underground does not produce sedimentary rocks. Most igneous rocks are produced by this process.

When molten rocks underground called magma is solidified in the subsurface, it results into the formation of igneous bodies.

  • Sedimentary rocks forms by the accumulation of sediments.
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4 0
3 years ago
A block slides forward 4.73 m while being pulled back by a 39.7 N force at 142° from the direction of motion. How much work does
lubasha [3.4K]

Answer:

work done by the force, W = -147.97 J

Given:

displacement of the box, s = 4.73 m

angle between force and displacement, \theta = 142^{\circ}

Force, F = 39.7 N

Solution:

Work done by the force can be expressed as the dot product of Force, F and displacement 's' is given by:

W = Fscos\theta

W = 39.7\times 4.73\times cos142^{\circ}

W = -147.97 J

where negative sign indicates that work is done on the block

6 0
4 years ago
Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 g of carbon-14, how much carbon-14 m
MakcuM [25]

Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

Explanation:

Half-life of sample of carbon -14= 5,730 days

\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5,730 days}=0.00012 day^{-1}

Let the sample present 11,430 years(t) ago = N_o

Sample left till today ,N= 0.060 g

N=N_o\times e^{-\lambda t}

ln[N]=ln[N]_o-\lambda t

\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days

\log[N_o]=1.9369

N_o=86.47 g

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

5 0
3 years ago
Read 2 more answers
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