An 18.0-Newton wooden block slides across a horizontal wooden floor at constant velocity.

An object with mass 80 kg moved in outer space. When it was at location <7, -34, -7> its speed was 14.0 m/s. A single cons

Sergio [31]

Answer:

W = -2080 J

Explanation:

initial position vector of the object is given as

$r_i = 7\hat i - 34\hat j - 7 \hat k$

similarly final position vector is given as

$r_f = 12\hat i - 42\hat j - 11\hat k$

now the displacement of the object is given as

$\vec d = \vec r_f - \vec r_i$

now we will have

$\vec d = (12\hat i - 42\hat j - 11\hat k) - (7\hat i - 34\hat j - 7 \hat k)$

$\vec d = 5\hat i - 8\hat j- 4\hat k$

now the force on the object is given as

$\vec F = (200\hat i + 460 \hat j - 150 \hat k)$

so here in order to find the work done

$W = \vec F . \vec d$

$W = (200\hat i + 460 \hat j - 150 \hat k). (5\hat i - 8\hat j- 4\hat k)$

$W = 1000 - 3680 + 600 = -2080 J$

7 0

3 years ago

)consider two positively charged particles, one of charge q0 (particle 0) fixed at the origin, and another of charge q1 (particl

I am Lyosha [343]

here force between two particles is given by Coulomb's law where two particles will either repel or attract each other

this force of attraction or repulsion is given by the equation

$F = \frac{kq_1 q_2}{r^2}$

here we know that

$q_1 , q_2$ = two point charges

r = distance between two charges

so now we will substitute all values in it

$F = \frac{kq_0q_1}{d_1^2}$

so above is the force between two point charges

5 0

3 years ago

How is sound different from wind?

Mashutka [201]

Answer:

Wind - the perceptible natural movement of the air, especially in the form of a current of air blowing from a particular direction. Sound wave - a wave of compression and rarefaction, by which sound is propagated in an elastic medium such as air.

Sound wave travels faster in the air when it is with the wind. ... In contrast, when sound wave travels against the wind, that is, when it travels in the opposite direction with the wind, the speed of sound will be reduced by the wind speed, resulting in a lower speed in the upper region.

Explanation:

please mark me brainliest

3 0

3 years ago

If a car travels 400m in 2o seconds how fast is it going ?

il63 [147K]

Assuming that what you typed meant 20 seconds, rather than 2o seconds, the car is travelling at a velocity of 200 meters per second.

6 0

3 years ago

two 100 kg astronauts are floating in space. the first astronaut is moving at 5 m/s while the second is at rest. the two astrona

Nata [24]

Answer:

The true statement is;

Neither momentum or kinetic energy is conserved

Explanation:

The question relates to the verification of the conservation of linear momentum, and kinetic energy

The given parameters are;

The mass of each astronaut = 100 kg

From which, we have;

The mass of the moving astronaut, m₁ = 100 kg

The mass of the stationary astronaut, m₂ = 100 kg

The initial velocity of the moving astronaut, v₁ = 5 m/s

The initial velocity of the stationary astronaut, v₂ = 0 m/s

The final velocity of both astronauts, v₃ = 3 m/s

The sum of the initial momentum of both astronauts is given as follows;

$P_{initial}$ = m₁·v₁ + m₂·v₂ = 100 kg × 5 m/s + 100 kg × 0 m/s = 500 kg·m/s

$P_{initial}$ = 500 kg·m/s

The sum of the final momentum of the astronauts is given as follows;

$P_{final}$ = m₁·v₃ + m₂·v₃ = (m₁ + m₂) × v₃ = (100 kg + 100 kg) × 3 m/s = 600 kg·m/s

$P_{final}$ = 600 kg·m/s

∴ $P_{initial}$ = 500 kg·m/s ≠ $P_{final}$ = 600 kg·m/s

$P_{initial}$ < $P_{final}$ , therefore, the sum of the linear momentum of both astronauts is not conserved

The sum of the initial kinetic energy of each astronaut is given as follows;

$K.E._{initial}$ = 1/2·m₁·v₁² + 1/2·m₂·v₂² = 1/2 × 100 kg × (5 m/s)² + 1/2 × 100 kg × (0 m/s)² = 1250 Joules

$K.E._{initial}$ = 1250 Joules

The sum of the final kinetic energy of the astronaut is given as follows;

$K.E._{final}$ = 1/2·m₁·v₃² + 1/2·m₂·v₃² = 1/2 × 100 kg × (3 m/s)² + 1/2 × 100 kg × (3 m/s)² = 900 joules

$K.E._{final}$ = 900 joules

$K.E._{initial}$ > $K.E._{final}$ , therefore, the kinetic energy is not conserved

From which we get that neither momentum or kinetic energy is conserved.

5 0

3 years ago