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Vesnalui [34]
4 years ago
12

How much energy is required to raise the temperature of one kilogram (liter) of water 1°c?

Physics
1 answer:
Gnoma [55]4 years ago
4 0

Heat required in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
Heat = 1 kg (4.18 kJ / kg C)( 1 C)
<span>Heat = 4.18 kJ energy needed</span></span>

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Which electromagnetic waves can have wavelengths of 300 nm?
Gelneren [198K]
I believe that the answer is A
6 0
3 years ago
What is necessary for fossil fuels to form? oxygen forming continuously for a long period of time sunlight shining continuously
Oksanka [162]
Heat and pressure acting over a long period of time
4 0
4 years ago
Read 2 more answers
A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values
Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or \lambda=20.0cm

c. The period, or T, of a wave is found in the equation

f=\frac{1}{T} were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

50.0=\frac{1}{T} so

T=\frac{1}{50.0} and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

f=\frac{v}{\lambda} and since we already have the frequency and we solved for the wavelength already, filling in:

50.0=\frac{v}{20.0} and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0
3 years ago
What is the magnitude of velocity for a 2000 kg car possessing 3000 kgm/s of momentum
Alex

For this case we have that by definition, the momentum is given by:

p = mv

Where,

  • <em>m: mass </em>
  • <em>v: speed </em>

Therefore, replacing values we have:

3000 = (2000) v

From here, we clear the value of the speed:

v = \frac {3000} {2000}\\v = 1.5 \frac {m} {s}

Answer:

The magnitude of velocity is:

v = 1.5 \frac {m} {s}

4 0
4 years ago
Read 2 more answers
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

7 0
3 years ago
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