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2 years ago

12

PLEASE HELP! How much heat is absorbed by 57g iron skillet when its temperature rises from 11°c to 30°C?

1 answer:

Amiraneli [1.4K]2 years ago

7 0

mass times specific heat times tem change.

57 x sh x (30-11)

convert to mks and look up sh iron

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the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i

Elanso [62]

Gravitational potential energy can be calculated using the formula:

$PE_{grav} =mgh$

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it.
$PE_{grav} =mgh$
$PE_{grav} =(10kg)(3.7m/s^{2})(1m)$
$PE_{grav} =37J$

7 0

2 years ago

Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

Bingel [31]

Answer:

$t=14.678\times 10^{-3}s$

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

$Q=Q_o(1-e^{-\frac{t}{RC}})$

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

or

$0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}$

or

$e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1$

taking log both sides, we get

$ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})$

or

$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

or

$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

or

$t=14.678\times 10^{-3}s$

3 0

2 years ago

Which of the following is NOT a function of MITOSIS?

DIA [1.3K]

Answer:

sexual reproduction

Explanation:

while sexual reproduction is a function of meiosis

6 0

2 years ago

An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the

Korolek [52]

Efficiency = (useful output) / (input)

Efficiency = (35 J) / (125 J) = 0.28 = 28%

8 0

2 years ago

Read 2 more answers

In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of

Airida [17]

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

3 0

2 years ago