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2 years ago

15

Given a triangle with sides X = 6.35 cm and Y = 12.25 cm with an angle of 90 degrees between them, find the length of the hypote

nuse and the size of the other two angles.

1 answer:

never [62]2 years ago

4 0

Answer:

the hypotenuse = 13.78 cm

Ф = 27.44°

θ = 62.56°

explanation:

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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

1 year ago

When two objects are in contact, moving together, which of the following statements must be true? Choose all that apply. When tw

Setler [38]

Answer:

The objects must have the same acceleration and the objects must exert the same magnitude force on each other.

Explanation:

The objects must have the same weight: FALSE. This is not needed, any two object can move together in contact no matter their mass.

The objects must have the same acceleration: TRUE. If they have different accelerations, they will separate since the distance each of them travel at a given time will be different.

The objects must have the same net force acting on them: FALSE. This is not needed, since what matters is acceleration, and a=F/m, so if both objects have different net force acting on them, they could have different masses also to compensate and result in the same acceleration.

The objects must exert the same magnitude force on each other: TRUE, this is the 3rd Newton Law, an action must follow the same reaction.

7 0

2 years ago

How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?

noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

$U=mgh$

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

$U=(50 kg)(9.8 m/s^2)(15 m)=7350 J$

3 0

1 year ago

The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change

mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

$\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})$

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m$

(b)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m$

(c)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m$

(d) The three lines belong to the ultraviolet region.

8 0

1 year ago