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earnstyle [38]
3 years ago
13

Why do we need to separate out certain things from a mixture

Physics
1 answer:
Nadusha1986 [10]3 years ago
7 0

To get the right solution of what we want

You might be interested in
Which of the following accelerations is the smallest in magnitude (ignore positive or negative signs)?
Basile [38]

The <em><u>gravity on the Earth</u></em> has acceleration is smallest in magnitude

Given:

  • Acceleration from 0 to 30 m/s in 2 s
  • Deceleration from 30 m/s to 10 m/s in 1.5 s
  • Gravity on Earth

To find:

The accelerations with the smallest magnitude

Solution:

  • Acceleration from 0 to 30 m/s in 2 s The acceleration in 2 seconds:

=\frac{30 m/s-0 m/s}{2s}=15 m/s^2

The magnitude of the acceleration is 15.

  • Deceleration from 30 m/s to 10 m/s in 1.5 s The deceleration in 1.5 seconds:

=\frac{10m/s-30 m/s}{1.5 s}=-13.4 m/s^2

The magnitude of the deceleration is 13.4.

  • Gravity on earth

The value of the acceleration due to gravity = g = 9.8 m/s^2

The magnitude of the acceleration is 9.8.

9.8 < 13.4 < 15

The <em><u>gravity on the Earth</u></em> has acceleration is smallest in magnitude.

Learn more about the acceleration here:

brainly.com/question/2437624?referrer=searchResults

brainly.com/question/7488209?referrer=searchResults

5 0
1 year ago
Bruce (the dog) is pulling on his lead with a force of 40N at an angle of 26º below the horizontal. Show that the horizontal com
saw5 [17]
You will use the equation for work which is w= F times distance so you will take 40N times 36N which equals 1440N
6 0
3 years ago
Read 2 more answers
A 0.0140-kg bullet is fired straight up at a falling wooden block that has a mass of 2.42 kg. The bullet has a speed of 555 m/s
7nadin3 [17]

Answer:

0.16 s

Explanation:

• Falling from rest (V_block= 0 m/s) the block attains a final velocity V_block before colliding with the bullet. This velocity is given by Equation 2.4 as

V_block(final velocity of block just before hitting) =V_0,block +at

where a is the acceleration due to gravity (a = —9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is

V_block = at

• During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that

(M_bullet+M_block)V_f = M_bullet*V_bullet+ M_block*V_block  

Total linear momentum after collision = Total linear momentum  before collision

Here V_f is the final velocity of the bullet/block system after the collision, and V_bullet and V_block  are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that V_f, the final velocity of the bullet/block system after the collision must have the same magnitude as V_block, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that of V_f =-V_block, Substituting this relation and Equation (1) into Equation (2) gives

(M_bullet + M_block)(-at) = M_bullet*V_bullet + M_block(at)

t = -M_bullet*V_bullet/a(M_bullet +2M_block)

 =-(0.0140-kg)*555 m/s/-9.8(0.0140-kg+2(2.42 kg)

 =0.16 s

3 0
2 years ago
A sled slides down a hill, reaches the level surface, and eventually comes to a stop. Which fact ultimately explains why the sle
Kitty [74]

Answer:

B. The presence of an unbalanced force(e.g friction) causes a moving object to stop.

Explanation:

As the friction is that force that can stop the sled upon reaching the levelled surface so the option b is correct.

7 0
2 years ago
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
2 years ago
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