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3 years ago

Adding resistors in series changes the total resistance of a circuit by

Physics

1 answer:

skelet666 [1.2K]3 years ago

6 0

Answer:

Increasing the resistance

Explanation:

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If an 83.00 g sample of Iron has a starting temperature of 297K and an ending temperature of 329K how much heat will be lost fro

diamong [38]

Question 1 :

m = mass of iron sample = 83

c = specific heat of iron sample = 0.450 J/g ⁰C

T₀ = initial temperature = 297 K

T = final temperature = 329 K

Q = heat lost

Heat lost is given as

Q = m c (T - T₀)

inserting the values

Q = (83) (0.450) (329 - 297) = 1195.2 J

Question 2 :

m = mass of sample = 59 g

c = specific heat of sample = ?

ΔT = change in temperature = 85 c

c = specific heat

Q = heat absorbed = 4814.4 J

Heat absorbed is given as

Q = m c

inserting the values

4814.4 = (59) (85) c

c = 0.96 J/g C

the sample is cholorofom

Question 3 :

m = mass of sample = 55 g

c = specific heat of sample = ?

T₀ = initial temperature = 78 c

T = final temperature = 95 c

Q = heat absorbed = 3908.3 J

Heat absorbed is given as

Q = m c (T - T₀)

inserting the values

3908.3 = (55) (95 - 78) c

c = 4.2 J/g C

the sample is water

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3 years ago

How many significant figures are in the measurement 230kg

Free_Kalibri [48]

There are 39 sig. figs.

6 0

3 years ago

Which one of the following scenarios accurately describes a condition in which resonance can occur?

oksian1 [2.3K]

I believe it's B. A vibrating tuning fork is struck and begins to vibrate as the object used to strike it is placed away from the tuning fork.

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3 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

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

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

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3 years ago

Please answer these few questions. This is supposed to be a science question but I had to put it as physics.

Kaylis [27]

I don’t understand how I’m supposed to answer that if you have to watch a video and answer certain questions on certain pages which we don’t have

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3 years ago

Read 2 more answers