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2 years ago

12

Two men are trying to carry a wooden pole. If one of them is weaker than other, how can they carry the pole hence making small l

oad for the weak man?

2 answers:

shepuryov [24]2 years ago

7 0

The stronger man can go more into the middle of the pole and the weak one can take the shorter end

Romashka-Z-Leto [24]2 years ago

6 0

Answer: Answer down below.

Explanation:

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EMERGENCY! PLS HELP

Karolina [17]

Answer:

0.75 grams

Explanation:

Today we have 6 grams

In 30 years, 3 grams remain

In 60 years 1.5 grams remain

in 90 years 0.75 grams remain

mathematically It would look like this

m = 6(0.5⁽ⁿ/³⁰⁾) where n is number of years

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3 years ago

A 2-kg object is moving 6 m/s in a horizontal direction.

german

Answer: A) O N

Explanation:

An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.

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3 years ago

A wave is travelling at 3000 m/s has a wavelength of 1 m.

Levart [38]

Answer:

a] 3000hz

b]3.33 × 10⁻⁴

ci]300

ii] 3000

iii]60,000

Explanation:

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3 years ago

What continent is located at 40 degrees north and 100 degrees west?

Rina8888 [55]

The continent located 40n and 100w is North America

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3 years ago

Read 2 more answers

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago