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aivan3 [116]
3 years ago
6

Calculate the area of a square with a length of 5cm

Physics
2 answers:
mihalych1998 [28]3 years ago
6 0

Answer:

25cm^2

Explanation:

area of square = side × side

length of side given = 5

area of this square = 5× 5

= 25cm^2

hope it helps

Mamont248 [21]3 years ago
4 0

Answer:

25cm^2

Explanation:

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The diagram shows a ramp with a toy car at the bottom. A string attached to the front of the car and the string goes over a pull
Annette [7]

The answer would be Gravity.

Gravity is pulling the weight down, which is pulling the car up the ramp.

3 0
4 years ago
A 285-mL flask contains pure helium at a pressure of 750 torr . A second flask with a volume of 455 mL contains pure argon at a
zaharov [31]

Answer:

P_{He}^|=288.85torr

Explanation:

Given data

P_{He}=750 torr\\V_{He}=285mL\\P_{Ar}=732 torr\\V_{Ar}=455 mL

Required

Partial pressure of helium

Solution

First calculate the total volume of gas mixture once the stopcock is opened

So

V_{total}=V_{He}+V_{Ar}\\V_{total}=285mL+455mL\\V_{total}=740mL

As temperature remains constant,by Boyle's Law we calculate the partial pressure of the helium gas

So

P_{He}V_{He}=P_{He}^|V_{total}\\P_{He}^|=\frac{P_{He}V_{He}}{V_{total}}\\P_{He}^|=\frac{750torr*285mL}{740mL}\\  P_{He}^|=288.85torr

5 0
3 years ago
Read 2 more answers
A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla
Scorpion4ik [409]

Answer:

 i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

    2 L = n λ           n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

     v = λ f

     λ = v / f

Let's replace in the first equation

    2 L = 1 (v / f₁)

For the shortest length L = L-l

   2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

    v = 2L f₁

    v = 2 (L-l) f₂

    2L f₁ = 2 (L-l) f₂

    L- l = L f₁ / f₂

    l = L - L f₁ / f₂

    l = L (1- f₁ / f₂)

.

Let's calculate

    l / L = (1- 309/463)

    i / L = 0.3326

4 0
3 years ago
Weight is influenced by which of the following
nignag [31]

Out of the given options, weight is influenced by mass and gravity

Answer: Option A

<u>Explanation: </u>

The object's mass is defined as the quantity of a matter with which the object is formed. It can change its state of matter but the quantity will remain the same. However, the weight is defined as how much force gravity exerts on the object's mass to pull it.

The mass is always same irrespective the location but the weight may vary from one place to the other while talking for the bigger picture. For example, the object's weight may be 60 kg on Earth but when it is measured on the moon, it will be lesser.

The weight of an object generally has nothing doing with the volume and it doesn't depend solely on the gravitational pull. The mass plays a crucial role.

                                  W=F=m \times g

6 0
3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
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