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2 years ago

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Density of water is 1000kg/m3. Find out the amount of mass of 5m3 water

1 answer:

Licemer1 [7]2 years ago

4 0

Answer:

5000 m³

Explanation:

density = mass / volume

mass = density × volume

mass = 1000 × 5

mass = 5000 ( kg )

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Gravity is a force. Gravity is also an example of __________________. . . A.. a theory. . B.. an established procedure. . C.. a

MissTica

The correct answer to this question is C - Gravity is a force. Gravity is also an example of a universal law. Well, according to Isaac Newton, anyway. According to Newton's Law of Universal Gravitation, 'every point mass attracts every single point mass by a force pointing along the line intersecting both paths.'
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2 years ago

A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th

GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

$W_g = m_A g(h - h_A)$

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

$W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})$

$W_f = -\mu m_A g cot\theta(h + h_A)$

Total work done on it

$W = m_A g(h - h_A) - \mu m_A g cot\theta(h + h_A)$

So answer will be

None of these

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2 years ago

Read 2 more answers

An object travels 7.5 m/s toward the west . Under the influence of a constant net force of 5.2 kN, it comes to rest in 3.2 s. Wh

Softa [21]

Answer:

m = 2218.67 kg

Explanation:

It is given that,

Initial velocity, u = 7.5 m/s

Final speed of an object, v = 0 (at rest)

Force, F = 5.2 kN

Time, t = 3.2 s

We need to find the mass of the object. Force acting on an object is given by :

F = ma

m is mass, a is acceleration

$F=\dfrac{m(v-u)}{t}\\\\m=\dfrac{Ft}{v-u}\\\\m=\dfrac{5.2\times 10^3\times 3.2}{0-7.5}\\\\m=2218.67\ kg$

So, the mass of the object is 2218.67 kg

5 0

2 years ago

Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575

marin [14]

Answer:

Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

Explanation:

Let the speed of the wind = Vw,

Speed of the plane in still air = Vsa,

The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

The coming trip the wind behind = 1575mi/3.5hrs = 450

Write the motion in equation form

First trip ( the plane flew into the wind)

Vaverage = Vsa - Vw

350 = Vsa - Vw

Second trip the wind was behind

450 = Vsa +Vw

Adding the two equation

800 = 2Vas

Vas = 800/2 = 400mi/hr

Substitute for Vas into equation 1

350mi/hr = 400mi/hr - Vw

Vw = 400-350 = 50mi/hr

6 0

2 years ago

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

jeyben [28]

Initial speed of Gazelle is along x direction and its value will be

$v_x = 5.6 m/s$

also its initial height is given as

$y = 2.5 m$

Part a)

now from kinematics along Y direction

$\Delta y = v_y t + \frac{1}{2} at^2$

as we know that

$\Delta y = 0$

$v_y = 0$

$a = 9.8 m/s^2$

$2.5 = 0 + \frac{1}{2} (9.8) t^2$

$t = 0.714 s$

Part b)

distance moved horizontally

$\Delta x = v_x t$

as we know that

$v_x = 5.6 m/s$

now we will have

$v_x = 5.6 (0.714) = 4m$

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

$v_{fy} = v_y + at$

$v_{fy} = 0 + (9.8)(0.714) = 7 m/s$

$v_x = 5.6 m/s$

so net speed will be

$v^2 = v_x^2 + v_y^2$

$v^2 = 7^2 + 5.6^2$

$v = 8.96 m/s$

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2 years ago