Can life be found/ sustain on Mars or other planets? What conditions are required to support life on another planet?

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o

Masja [62]

Answers is F=7.84 N

Friction force resists the effect of horizontal force and trying to approch to a limiting force.

we have formula for limiting friction force between block and floor

F=Ц N

where N=mg

putting values we get answer.

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3 0

2 years ago

A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 =50m/s hits an iron bloc

Pavel [41]

Answer: 0.001 m

Explanation:

In order to get the máximum height reached by the iron block, we can use the energy conservation principle, as all the kinetic energy impressed upon the iron blck by the rubber bullet during the collision, becomes gravitational potential energy, as follows:

½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

P1 = mb . vb = 0.025 kg. 50 m/s = 1.25 kg. m/s

The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

Now, we can replace this value in the equation (1) above:

½ . 15 Kg. (0.14)2 (m/s)2 = 15 Kg. 9.8 m/s2. H

Solving for H, we have:

H = 0.001 m

3 0

2 years ago

What is a Non-Example of Newton's 1st law?

guajiro [1.7K]

Answer:

Say a 14 year old girl was at a construction site and she was asked to move something like a 10,000 pound brick( one brick). She would be acting on it as the unbalanced force but they would still not change their position.

so to say the girl would be doing everything she could to move that brick but the brick would still be in that same spot so the unbalanced force (the girl) would be acting on the thing that was at rest but it wouldn't move.

so the unbalanced force would not really be acting on the thing at rest; even though the unbalanced force was doing something to the brick.

( just think about it and you will eventually get it...just imagine in your head...)

Explanation:

5 0

1 year ago

HELPPpPpPpPpPp!!!! ASAP !!!

DENIUS [597]

Answer:

50 N

Explanation:

8 0

1 year ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

b)

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

6 0

2 years ago