We're adding two vectors here. The first is 300 Newtons to the right, which we can write as (300, 0), meaning 300 to the right, 0 up.
The second is 300 at let's say a 45 degree angle down. For the components we have an isosceles right triangle with hypotenuse 300, so the components are both magnitude 300/√2 = 150√2. So we can write this vector (150√2, -150√2), the negative sign because it points down in the y direction.
Adding is componentwise. The resulting force is (300+150√2, -150√2).
That has square magnitude
r² = (300+150√2)² + (-150√2)² = 150² ( (2+√2)² + (√2)² )
= 150²( (6 + 4√2) + 2)
= 300²(2+√2)
so
r = 300 √(2+√2) Newtons
That's the answer; I'm not sure if your class expects a calculator approximation, which is 554.3 Newtons.
Answer:
A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle. It has two circular objects which includes a larger disc and a small cylinder both joined at the center.
Explanation:
Answer:
The temperature of the CMB is cooler, not hotter, than at the time of the big bang.
Explanation:
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A :-) for this question , we should apply
F = ma
Given - F = 12 N
a = 0.20 m/s^2
Solution -
F = ma
12 = m x 0.20
m = 12 by 0.20
m = 60 kg
.:. The mass is 60 kg.
Answer:
a.) The main scale reading is 10.2cm
b.) Division 7 = 0.07
c.) 10.27 cm
d.) 10.31 cm
e.) 10.24 cm
Explanation:
The figure depicts a vernier caliper readings
a.) The main scale reading is 10.2 cm
The reading before the vernier scale
b.) Division 7 = 0.07
the point where the main scale and vernier scale meet
c.) The observed readings is
10.2 + 0.07 = 10.27 cm
d.) If the instrument has a positive zero error of 4 division
correct reading = 10.27 + 0.04 = 10.31cm
e.) If the instrument has a negative zero error of 3 division
correct reading = 10.27 - 0.03 = 10.24cm
