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solong [7]
2 years ago
15

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to

an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of 256m/s
Physics
1 answer:
Salsk061 [2.6K]2 years ago
7 0

Explanation:

1.2 \mathrm{N} ; 2 \mathrm{N}

2.200 \mathrm{N} ; 200 \mathrm{N}

4.2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}

5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}

6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}

8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of 256 \mathrm{m} / \mathrm{s}

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

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...the potential energy that you build while going up the hill on the roller coaster could be let go as kinetic energy -- the energy of motion that takes you down the hill of the roller coaster.
4 0
2 years ago
A small crack occurs at the base of a 15.0-m-high dam. The effective area through which water leaves is 2.30 × 10-3 m2. (a) Igno
vova2212 [387]

Answer

given,                                              

height of the dam = 15 m            

effective area of water = 2.3 x 10⁻³ m²

Using energy conservation              

    m g h = \dfrac{1}{2}mv^2

    v= \sqrt{2gh}                  

    v= \sqrt{2\times 9.8 \times 15}

    v= \sqrt{294}              

           v = 17.15 m/s            

 discharge of water

      Q = A V                            

      Q = 2.3 x 10⁻³ x 17.15    

      Q = 0.039 m³/s

3 0
2 years ago
A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the
Elis [28]

Answer:

a)R = 171μΩ

b)E = 1.7 *10^{-4} V/m

c)R_{2} = 1.16 *10^{-4}Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder  =  1.1 cm

resistivity depends on the distance x from the left

p(x)=a+bx^2 ............(i)

using equation

R = \frac{pl}{a}

let dR is the resistance of thickness dx

dR =\frac{p(x)dx}{a}

where p(x) is resistivity  l is length

a is area

\int\limits^R_0 {dR}  =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx  \\.........................(2)

after integration

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}  ...............(3)

it is given p(0) = a = 2.25 * 10 ^{-8}Ωm

p(L) = a + b(L)^2  = 8.5 * 10 ^{-8} Ωm

8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\

(here * stand for multiplication )

on solving we get

b = 2.78* 10^{-8} Ωm

put each value of a  and b and r value in equation 3rd we get

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}

R = 1.71 * 10^{-4}Ω

R = 171μΩ

FOR (b)

for mid point  x = L/2

E = p(x)L

for x = L/2

p(L/2) = a+b(L/2)^2

for given current  I = 1.75 A

so electric field

 

E = \frac{[a+b(L/2)^2]I }{\pi  r^2}

by substitute the values

we get;

E = 1.7 *10^{-4} V/m

(here * stand for multiplication )

c ).

75 cm means length will be half

 that is   x =  L/2

integrate  the second equation with upper limit  L/2  

Let resistance is R_{1}

so after integration we get

R_{1}  =  \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}

substitute the value of a , b and L we get

R_{1} = 5.47 * 10 ^{-5}Ω

for second half resistance

R_{2} =  R- R_{1}

R_{2}  = 1.7 *10^{-4} -5.47 *10^{-5}

R_{2} = 1.16 *10^{-4}Ω

(here * stand for multiplication )

5 0
2 years ago
Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an
vodomira [7]

Answer:

A sample of 5.2 mg  decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510  requires about 9  half-lives which is 30 * 9 = 270 yrs

7 0
2 years ago
A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
scZoUnD [109]
Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by  50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

8 0
2 years ago
Read 2 more answers
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