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Tanzania [10]
3 years ago
8

What is the energy in space left over from the big bang called

Physics
1 answer:
Lilit [14]3 years ago
5 0
It is called the CMBR, which stands for cosmic microwave background radiation. It was discovered by Arno Penzias and Robert Wilson in 1964.
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What is meant by resistance and voltmeter?​
LUCKY_DIMON [66]

Answer:

<em>The internal resistance of an ideal ammeter will be zero since it should allow current to pass through it. Voltmeter measures the potential difference, it is connected in parallel. .</em>

Explanation:

<h3><em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h3>
7 0
2 years ago
An object has a mass of 30 grams and measures 3cmx2cmx1cm. What is the density of the object?
dimulka [17.4K]
The density of the object is the ratio of its mass and volume. From the given dimensions above, we determine the volume through the equation,

      V = L x W x H

Substituting,

    V = (3 cm)(2 cm)(1 cm) = 6 cm³

From the idea presented above,
    d = m/V

Substituting the known values,

   d = (30 g)/ (6 cm³) = 5 g/cm³

ANSWER: 5 g/cm³
4 0
3 years ago
The density of an object is dependent upon the object’s mass and ---
kotegsom [21]

Answer:Volume

Explanation:

Density = mass/ Volume

7 0
3 years ago
At what speed will a box be falling at a time t = 0.75 s after being dropped?
IRINA_888 [86]

Explanation:

Initial speed(u)= 0 m/s (Ball is dropped)

time(t)= 0.75 s

acceleration(a)= 10 m/s² (gravity)

Final speed(v)= u+at

v=0+(10)× 0.75

v=7.5 m/s

Speed is 7.5 m/s

8 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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