Answer:
B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.
Explanation:
A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.
In the motion map:
- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration
- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed
In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.
Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
Explanation:
a. Net force is mass times acceleration (Newton's second law).
∑F = ma
∑F = (5.0 kg) (2.0 m/s²)
∑F = 10 N
b. The net force is the sum of the individual forces.
10 N = F − 5 N
F = 15 N
c. Friction force here is mgμ.
mgμ = 5 N
(5.0 kg) (10 m/s) μ = 5 N
μ = 0.1
