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tatiyna
3 years ago
8

Two forces,one of 12 N and another of 24 N,act on a body in such a way that they make an angle of 90degree with each other.Find

the resultant of the two forces
Physics
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

26.83 N.

Explanation:

If the angle between two vector is 90°, to get the resultant, we use Pythagoras theorem.

a² = b²+c²......................... Equation 1

Where a = R = Resultant, b =  12 N, c = 24 N.

Substitute these values into equation 1

R² = 12²+24²

R² = 144+576

R² = 720

√R² = √720

R = 26.83 N.

Hence, the result of the two force is 26.83 N.

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Distance is a vector quantity ... true or false
Elena-2011 [213]
Answer:False
Reason:
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3 years ago
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Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
3 years ago
A manometer using oil (density 0.900 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases
Mila [183]

Answer:

Rise in level of fluid is 0.11 m

Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm

Solution:

As per the question:

Density of oil, \rho_{o} = 0.900\ g/cm^{3} = 900\ kg/m^{3}

Change in Pressure in the tank, \Delta P = 7.28\ mmHg

Density of the mercury, \rho_{m} = 13.6\ g/cm^{3} = 13600\ kg/m^{3}

Now,

To calculate the rise in the level of fluid inside the manometer:

We know that:

1 mmHg = 133.332 Pa

Thus

\Delta P = 7.28\ times 133.332 = 970.656\ Pa

Also,

\Delta P = \rho_{o} gh

where

g = acceleration due to gravity

h = height of the fluid level

970.656 = 900\times 9.8\times h

h = 0.11 m

Now, if mercury is used:

\Delta P = \rho_{m} gh

970.656 = 13600\times 9.8\times h

h = 0.00728 m = 7.28 mm

3 0
4 years ago
A 33.0 g iron rod, initially at 22.7 ∘c, is submerged into an unknown mass of water at 63.3 ∘c, in an insulated container. the f
professor190 [17]
When the iron and the water reach thermal equilibrium, they have same temperature, T=58.5^{\circ}C.
We can consider this as an isolated system, so the heat released by the water is equal to the heat absorbed by the iron.

The hear released by the water is:
Q_w =m_w C_{sw} \Delta T_w
where m_w is the water mass, C_{sw}=4.186 J/(g^{\circ}C) is the specific heat of the water, and \Delta T_w = 63.6^{\circ}-58.5^{\circ}=5.1^{\circ}C is the variation of temperature of the water.

Similarly, the heat absorbed by the iron is:
Q_i = m_i C_{si} \Delta T_i
where m_i = 33.0 g is the iron mass, C_{si}=0.444 J/(g^{\circ}C) is the iron specific heat, and \Delta T_i = 58.5^{\circ}-22.7^{\circ}=35.8^{\circ}C is the variation of temperature of the iron.

Writing Q_w=Q_i and replacing the numbers, we can solve to find mw, the mass of the water:
m_w= \frac{m_i C_{si} \Delta T_i}{C_{sw} \Delta T_w} =24.6 g
4 0
3 years ago
At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc
nasty-shy [4]

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

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I = √10.2

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The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

4 0
3 years ago
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