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4 years ago

How much heat is absorbed by a 27g iron skillet when its temperature rises from 13oC to 23oC?

1 answer:

8 0

The equation for heat would be q = m*Delta T* S

Where q = the heat absorbed in joules
m = mass
Delta T = change in temperature
s = specific heat capacity of the material

for iron s = 0.450

Delta T = 23 - 13 = 10 degrees Celsius

m is given as 27 grams.

q = 27 * 10 * 0.450 = 121.5 joules

You may need to round the answer depending on what the teacher is looking for.

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A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0

3 years ago

Read 2 more answers

dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

7 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago

5. No matter what month and date your birthday is on, NASA says it has a significant picture from

padilas [110]

Answer:

The Hubble space telescope.

Explanation:

Hubble is a telescope that observers the sky 24/7 non-stop, which means that for every day of the year it would have made a significant discovery, which of course includes your birthday. Furthermore, you can actually go to NASA website and find out what discovery was made on your birthday! This shows both the vastness of the universe <em>(it really has to be huge for a telescope to have a discovery for each day of the year!) </em> and the ceaseless work of the telescope!

7 0

3 years ago

250 mL 0.1 M HCl solution is mixed with 250 mL

pishuonlain [190]

Answer:

The concentration of OH⁻ in the mixture is 0.05 M

Explanation:

The reaction of neutralization between HCl and NaOH is the following:

H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)

The number of moles of HCl is:

$n_{HCl} = C*V = 0.1 mol/L*0.250L = 0.025 moles$