Answer:
μ = 0.350
Explanation:
For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:
In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):
Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.
A bathroom scales works due to gravity. Under normal
conditions, a reading can be obtained when your body is pushing some force on
the scale. However in this case, since you and the scale are both moving
downwards, so your body is no longer pushing on the scale. Therefore the answer
is:
<span>The reading will drop to 0 instantly</span>
Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
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:
Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
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