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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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Please help will make brainliest

kumpel [21]

The Gulf of Mexico
OR
near trenches, in the middle of the ocean, the continental shelf, and lastly, in the United States

3 0

3 years ago

Read 2 more answers

Titanium is a metal used to make golf clubs. A rectangular bar of this metal measuring 1.96 cm x 2.19 cm x 2.63 cm was found to

masha68 [24]

$4.012\frac{gr}{cm^3}$

Explanation

the density of an object is given by:

$\text{Density(d)}=\frac{mass(m)}{\text{volume(v)}}$

Step 1

find the volume of the bar

a)find the volume of the rectangular bar.

the volume of a rectangular prism is given by:

$\text{Volume}=\text{ length}\cdot widht\cdot depth$

replace

$\begin{gathered} \text{Volume}=(\text{ 2.63}\cdot2.19\cdot1.96)(cm^3) \\ \text{Volume}=11.289012(cm^3) \end{gathered}$

Step 2

now,

Let

$\begin{gathered} \text{Volume}=11.289012(cm^3) \\ \text{mass}=\text{ 45.3 gr} \end{gathered}$

replace in the formula

$\begin{gathered} \text{Density(d)}=\frac{mass(m)}{\text{volume(v)}} \\ d=\frac{45.3\text{ gr}}{11.289012(cm^3)} \\ d=4.012\frac{gr}{cm^3} \end{gathered}$

therefore, the answer is

$4.012\frac{gr}{cm^3}$

I hope this helps you

4 0

1 year ago

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

yawa3891 [41]

Answer:

Isothermal : P2 = ( P1V1 / V2 ) , work-done $pdv = nRT * In( \frac{V2}{v1} )$

Adiabatic : : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$ , work-done =

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

$pdv = nRT * In( \frac{V2}{v1} )$

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

$P1V1^y = P2V2^y$

where y = 5/3

hence : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$

Work-done

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Where $T2 = T1V1^(2/3)/V2^(2/3)$

3 0

3 years ago

When is the velocity of a mass on a spring at its maximum value?

ehidna [41]

Answer:

A. when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

$E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.

8 0

4 years ago

Read 2 more answers

The second-order bright fringe in a single-slit diffraction pattern is 1.35 mm from the center of the central maximum. The scree

Dafna11 [192]

Answer:

The wavelength is 754.2 nm.

Explanation:

Given that,

Diffraction pattern y= 1.35 mm

Width = 0.838 mm

Distance D= 75 cm

We need to calculate the wavelength

Using formula of diffraction pattern

$y=\dfrac{m\lambda D}{d}$

$\lambda=\dfrac{yd}{mD}$

Where, y = diffraction pattern

m = order

d = width

D = distance

Put the value into the formula

$\lambda=\dfrac{1.35\times10^{-3}\times0.838\times10^{-3}}{2\times75\times10^{-2}}$

$\lambda=7.542\times10^{-7}\ m$

$\lambda=754.2\ nm$

Hence, The wavelength is 754.2 nm.

4 0

3 years ago