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3 years ago

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Physics

1 answer:

Zinaida [17]3 years ago

4 0

I think it’s sleepy, but I’m not for sure

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A crane lifts a 1,750 kg mass using a steel cable whose mass per unit length is 0.88 kg/m. What is the speed of transverse waves

Sauron [17]

Answer:

139.6m/s

Explanation:

Calculate the tension first, T=m*g

mass(m): 1750kg, gravity(g): 9.8m/s^2

T= 1750*9.8

=17150N

Then calculate the wave speed using the equation v = √ (T/μ)

v= √(17150N)/(0.88kg/m)

=139.6m/s

4 0

3 years ago

an object is moving with a speed of 35 m/s and has a kinetic energy of 1500 J, what is the mass of the object?

Elena-2011 [213]

You'd use the equation kinetic energy=mass*0.5*speed^2
So you'd rearrange this to get mass =kinetic energy /0.5 *speed^2
Which is mass= 1500J/0.5*35^2
=2.44897959183673469........kg

5 0

3 years ago

The latent heat of vaporization for water at room temperature is 2430 J/g.

alukav5142 [94]

Answer:

1) $kinectic energy=7.26*10^-^2^0J$

2) $V= 2.0m/s$

3) $T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place

Explanation:

From the question we are told that

latent heat of vaporization for water at room temperature is 2430 J/g.

1)Generally in determining the molar mass of water evaporated we have that

-One mole (6.02 x 10. 23 molecules)

-Molar mass of water is 18.02 g/mol

Mathematically the mass of water is give as

$M=\frac{18.02}{6.02*10^-^2^6}$

$M=3*10^-^2^3g$

Therefore

$kinectic energy=2430J/g*3*10^-^2^3g$

$kinectic energy=7.26*10^-^2^0J$

b)Generally the evaporation speed V is given as $V= \sqrt{\frac{K.E*2}{m} }$

Mathematically derived from the equation

$\frac{1}{2} mv^2 =K.E$

To Give

$V= \sqrt{\frac{K.E*2}{m} }$

$V= \sqrt{\frac{7.26*10^-^2^0J*2}{3*10^-^2^3g} }$

$V= 2.0m/s$

c)Generally the equation for velocity $Vrms=\sqrt{\frac{3RT}{M} }$

Therefore

Effective temperature T is given by

$T=\frac{\sqrt{v}*m}{R}$

where

$T=\frac{\sqrt{2.0m/s}*6.02*10^-^2^6}{0.082057 L atm mol-1K-1}$

$T=3.5*10^3K$

4) The Molecules do not burn because of the presences of hydrogen bond in place

3 0

3 years ago

An 85 kg object is moving at a constant speed of 15 m/s in a circular path, which has a radius of 20 meters. What centripetal fo

UNO [17]

As we know that centripetal force =mv^2/r
given data is
m = mass
v = speed
r = radius
putting values we get

= 85 x 15^2 / 20

= 956.25 N
option d is correct

7 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$