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sleet_krkn [62]
3 years ago
5

A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two

vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers.
Physics
1 answer:
Andrew [12]3 years ago
4 0
A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers. <span>The given values are
F1 = 200 N</span> F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N
F2 = 400 - 200
F2 = 200 N



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Answer:

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Explanation:

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3 years ago
32. (FR) A mass moving at 25 m/s slides along a rough horizontal surface. The coefficient of friction is 0.30. (A) Use force and
scoray [572]

Answer:

A)s = 104.16 m

b)s= 104.16 m

Explanation:

Given that

u = 25 m/s

μ = 0.3

The friction force will act opposite to the direction of motion.

Fr= μ m g

Fr= -  m a

a=acceleration

μ m g = - m a

a= - μ g

a= - 0.3 x 10 m/s²          ( take g= 10 m/s²)

a= - 3  m/s²

The final speed of the mass is zero ,v= 0

We know that

v² = u² +2 a s

s=distance

0² = 25² - 2 x 3 x s

625 = 6 s

s = 104.16 m

By using energy conservation

Work done by all the forces =Change in the kinetic energy

- Fr.s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Negative sign because force act opposite to the displacement.

- \mu\ m\ g \ s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

-\mu\ g \ s=\dfrac{1}{2}v^2-\dfrac{1}{2}u^2

-0.3\times 10\times \ s=\dfrac{1}{2}\times 0^2-\dfrac{1}{2}\times 25^2

- 3 x 2 x s = - 625

s= 104.16 m

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3 years ago
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Active transport move molecules from low concentrations to high concentrations.
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3 years ago
Sound waves with frequency higher that 20,000 Hz are referred to as
Solnce55 [7]
<span>Sound waves with frequency higher that 20,000 Hz are referred to as "Ultrasonic"

Hope this helps!</span>
4 0
3 years ago
A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp
Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0
2 years ago
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