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Bumek [7]
3 years ago
11

At which point is the object not moving?

Physics
1 answer:
Alenkinab [10]3 years ago
8 0
It’s none of those because it’s moving at a constant rate
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A ball is dropped off the side of a bridge. after 1.55 s, how far has it fallen
givi [52]

Answer:

38.64 feet

Explanation:

x=x0 + vx0t + 1/2axt2

x= 0 +  0  + 1/2 X 32.17 ft/sec2 X 1.55 sec2

x = 38.64 feet

7 0
3 years ago
A​ right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 ​lb/ft3
tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ ft^{3} to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = \pi r^{2} =  \pi 4^{2} =16\pi ft^{2}

Now,

Work done required = \int_{0}^{8} 52\times 16\pi (21 - y)dy

                                  = 832\pi \int_{0}^{8} (21 - y)dy

                                  = 832\pi([ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\)

                                  = 113152\pi = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0
4 years ago
A rock is thrown straight downward from a tree limb with an initial velocity v0. The rock has constant downward acceleration of
viktelen [127]

Answer:

v₀ = 15 m/s

Explanation:

given,

initial velocity = v₀

down acceleration of rock = 10 m/s²

rock distance

  S₄ = 7 x S₁

From kinematic equations

S = v₀ t+0.5 at²

at t = 1 s

S₁ = v₀ (1)+0.5 x 10 x 1²

S₁ = v₀+ 5......(1)

at t = 4 s

 S₄ = v₀ (4)+0.5 x 10 x 4²

 S₄ = 4 v₀+80.....(2)

from equation (1) and  (2)  

7( v₀ + 5 ) = 4 v₀ +80

3 v₀ = 80 - 35

3 v₀ = 45

v₀ = 15 m/s

3 0
3 years ago
You place a chunk of naturally radioactive (it means not enriched for nuclear purposes) material on the not very exact scale and
Ber [7]

Answer:

he mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.

Explanation:

A radioactive material is transformed into another material by the emission of some particular radioactive ones, the most common being alpha and beta rays, which is why in the transformation process a certain amount of mass is lost. The process is described by the expression

             

              N = No e^{- \alpha /t}

 

From this expression the quantity half life time (T_{1/2}) is defined with time so that half of the atoms have been transformed

           

            T_{1/2} = ln 2 /λ

in this case it does not indicate that T_{1/2}= 20 days is worth, for which periods have passed, in the first the number of radioactive atoms was reduced to half the number, leaving N´ and the second halved the number of nuclei that they were radioactive, leaving radioactive nuclei

first time of life

              N´ = ½ N

second time of life

              N´´ = ½ N´

              N´´ = ¼ N

consequently in the sample at the end of these two decay periods we have, assuming that after each emission the atom is stable (non-radioactive). After the first emission there are n₁ = N / 2 stable atoms, after the second emission n₂ = ¼ N stable atoms are added and there are still n₃ = ¼ N radioactive atoms, so the total number of atoms is

 

             n_total = n₁ + n₂ + n₃

Recall that the mass of the initial radioactive atoms is m₁, when transforming its mass of stable atoms is m₂ where

            m₂ < m₁

therefore mass of

 

             m_total = m₂ N / 2 + m₂ N / 4 + m₁ N / 4

             m_total = m₂ ¾ N + m₁ ¼ N

             m_total = N (  ¾ m₂ + ¼ m₁)

Since the mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.

8 0
3 years ago
How much magnification is needed to see a nanometer?.
ArbitrLikvidat [17]

20,000,000



To see things at a nanometer, which is a trillionth of a meter, you would need to increase magnification nearly 20,000,000 times those are ionizing atoms
6 0
2 years ago
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