<span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:
Ff = UsN
where Us is the coefficient of static friction and N is the normal force.
In order to get the crate moving you must first apply enough force to overcome the static friction:
Fapplied = Ff
Since Fapplied = 43 Newtons:
Fapplied = Ff = 43 = UsN
and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11
43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
<span>1. It must be an object which independently orbits the Sun (this means moons can't be considered planets, since they orbit planets)
2. It must have enough mass that its own gravity pulls it into a spheroidal shape.
3. </span><span>It must be large enough to "dominate" its orbit (i.e. its mass must be much larger than anything else which crosses its orbit).</span>
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
Answer:
yes
Explanation:
Let's solve your equation step-by-step.
4x+3=−5
Step 1: Subtract 3 from both sides.
4x+3−3=−5−3
4x=−8
Step 2: Divide both sides by 4.
4x / 4 = −8 / 4
x=−2
Hope it helps,
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Thank you
Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L = ![\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%260%260%5C%5Cpx%26py%260%5Cend%7Barray%7D%5Cright%5D)
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
