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2 years ago

True or false. when objects collide , some momentum is lost

Physics

2 answers:

ycow [4]2 years ago

5 0

Answer:

It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

omeli [17]2 years ago

5 0

Answer: True

Explanation:

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If 300. mL of water are poured into the measuring cup, the volume reading is 10.1 oz . This indicates that 300. mL and 10.1 oz a

tigry1 [53]

Answer:

Milliliters to Ounces Conversions

some results rounded

mL - fl oz

200.00 6.7628

200.01 6.7631

200.02 6.7635

200.03 6.7638

200.04 6.7642

200.05 6.7645

200.06 6.7648

200.07 6.7652

200.08 6.7655

200.09 6.7658

200.10 6.7662

200.11 6.7665

200.12 6.7669

200.13 6.7672

200.14 6.7675

200.15 6.7679

200.16 6.7682

200.17 6.7686

200.18 6.7689

200.19 6.7692

200.20 6.7696

200.21 6.7699

200.22 6.7702

200.23 6.7706

200.24 6.7709

mL fl oz

200.25 6.7713

200.26 6.7716

200.27 6.7719

200.28 6.7723

200.29 6.7726

200.30 6.7729

200.31 6.7733

200.32 6.7736

200.33 6.7740

200.34 6.7743

200.35 6.7746

200.36 6.7750

200.37 6.7753

200.38 6.7757

200.39 6.7760

200.40 6.7763

200.41 6.7767

200.42 6.7770

200.43 6.7773

200.44 6.7777

200.45 6.7780

200.46 6.7784

200.47 6.7787

200.48 6.7790

200.49 6.7794

mL fl oz

200.50 6.7797

200.51 6.7800

200.52 6.7804

200.53 6.7807

200.54 6.7811

200.55 6.7814

200.56 6.7817

200.57 6.7821

200.58 6.7824

200.59 6.7828

200.60 6.7831

200.61 6.7834

200.62 6.7838

200.63 6.7841

200.64 6.7844

200.65 6.7848

200.66 6.7851

200.67 6.7855

200.68 6.7858

200.69 6.7861

200.70 6.7865

200.71 6.7868

200.72 6.7872

200.73 6.7875

200.74 6.7878

mL fl oz

200.75 6.7882

200.76 6.7885

200.77 6.7888

200.78 6.7892

200.79 6.7895

200.80 6.7899

200.81 6.7902

200.82 6.7905

200.83 6.7909

200.84 6.7912

200.85 6.7915

200.86 6.7919

200.87 6.7922

200.88 6.7926

200.89 6.7929

200.90 6.7932

200.91 6.7936

200.92 6.7939

200.93 6.7943

200.94 6.7946

200.95 6.7949

200.96 6.7953

200.97 6.7956

200.98 6.7959

200.99 6.7963

Explanation:

5 0

3 years ago

Read 2 more answers

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

2 years ago

Which of the following gases are typically used for colorful lighting when an electric current is apllied

Leona [35]

Well i really need to see the choices but i think you mean neon

8 0

3 years ago

Read 2 more answers

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

3 years ago

Read 2 more answers

A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg

dezoksy [38]

Answer:

For cast iron we have

$h = 0.92 m$

For copper

$h = 1.05 m$

For Lead

$h = 1.23 m$

For Zinc

$h = 2.43 m$

Explanation:

As we know that final speed of the block is calculated by work energy theorem

$W_f + W_g = \frac{1}{2}mv^2$

now we have

$-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2$

now we have

$v^2 = 2gh - 2\mu_k g h cot\theta$

$v = \sqrt{2gh(1 - \mu_k cot\theta)}$

For cast iron we have

$4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}$

$h = 0.92 m$

For copper

$4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}$

$h = 1.05 m$

For Lead

$4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}$

$h = 1.23 m$

For Zinc

$4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}$

$h = 2.43 m$

4 0

3 years ago