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To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

3 years ago

Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

$\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}$

here we know that

$\mu_1 = 1.473$

$h_o = 5 cm$

$\mu_2 = 1$

now by above formula

$\frac{1.473}{5} = \frac{1}{h_i}$

$h_i = 3.39 cm$

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0

3 years ago

The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in a medium. Which true statement i

Alika [10]

Answer:

A. Materials with a low index of refraction cause light to refract very little.

7 0

3 years ago

Read 2 more answers

A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.

Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is 1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

6 0

2 years ago

Question 2 (10 points)

kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

$a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2$

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

$F=ma$

where

m = 1390 kg is the mass of the truck

$a=-3.6 m/s^2$ is the acceleration

And substituting,

$F=(1390)(-3.6)=-5004 N$

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago