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kherson [118]
3 years ago
6

Find the acceleration of a body whose velocity increases from 11ms-1 to 33ms-1 in 10 seconds

Physics
2 answers:
Anvisha [2.4K]3 years ago
8 0
Acceleration is equivalent to the rate of change of velocity.

Calculate the difference of velocity, and divide by the total time it took for the change to happen.

(33 \frac{m}{s}  - 11 \frac{m}{s} ) \div 10seconds \\  = 2 \frac{m}{ {s}^{2} }
That is the answer.
Alex777 [14]3 years ago
7 0
As we know that 
a= dell   V/t
then put values
a=   2ms/^2
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Which of the following statements does NOT accurately describe the Berlin Airlift?
Karo-lina-s [1.5K]

The Berlin Airlift is best described as the aircraft used to delivered needed food and supplies to the city of Berlin through the air because all other routes were blocked by the Soviet Union.

<h3>What is Berlin Airlift?</h3>

The Berlin airlift was a 1940s military operation that supplied West Berlin with food and other vital goods by air after the Soviet Union blockaded the city.

Thus, the Berlin Airlift is best described as the aircraft used to delivered needed food and supplies to the city of Berlin through the air because all other routes were blocked by the Soviet Union.

Learn more about Berlin Airlift here: brainly.com/question/1104371

#SPJ1

8 0
2 years ago
Which element would you expect to gain one electron in a chemical reaction ?
Annette [7]

Answer: Hydrogen

Explanation: Which element would you expect to gain one electron in a chemical reaction ? HYDROGEN

Non-metals are more likely to gain electrons!

***If you found my answer helpful, please give me the brainliest, please give a nice rating, and the thanks ( heart icon :) ***

4 0
2 years ago
The earth has a mass ME = 5.98 · 1024 kg and the moon has a mass MM = 7.36 · 1022 kg. The distance from the center of the earth
ivann1987 [24]

Answer:

2 x 10^20 N

Explanation:

Me = 5.98 x 10^24 kg

Mm = 7.36 x 10^22 kg

r = 3.82 x 10^5 km = 3.82 x 10^8 m

The gravitational force between earth and moon is

F = G Me x Mm / r^2

F = (6.67 x 10^-11 x  5.98 x 10^24 x 7.36 x 10^22) / (3.82 x 10^8 x 3.82 x 10^8)

F = 2 x 10^20 N

6 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
If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

7 0
3 years ago
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