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2 years ago

100 POINTS PLEASE HELP

Physics

2 answers:

deff fn [24]2 years ago

8 0

Answer:

Explanation:

Spilling the force of 60N on the rope at an angle θ in x-y direction:

fx = Fcosθ = 60cosθ

fy = Fsinθ = 60sinθ

cosθ and sinθ both vary from 0 to 1 between 0° to 90° but they go in opposite direction:

cos0° = 1 and cos90° = 0

sin0° = 0 and sin90° = 1

sin45° = cos45° = 0.7071

When θ increases from 0° to 90°,

θ=0°, fx = 60cos0° = 60N, fy = 60sin0° = 0N

θ=90°, fx = 60cos90° = 0N, fy = 60sin90° = 60N

θ=45°, fx = 60cos45° = fy = 60sin45° = 42.43N

Pavlova-9 [17]2 years ago

6 0

Answer:

Explanation:

vector force F = 60N = Fx + Fy

where Fx = 60 cos(theta) and Fy = 60 sin(theta)

cos decreases from 1 to 0 from 0deg to 90 deg

sin increases from 0 to 1 from 0deg to 90deg

so Fx=60N and Fy=0N when theta=0deg

Fx=60cos45deg and Fy=60sin45deg when theta=45deg

Fx=0N and Fy=60N when theta=90deg

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The resistance (R) of a copper wire varies directly as its length (L). Write this relation as a formula using k as the constant

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Read 2 more answers

A model rocket is launched straight upward with an initial speed of 56.5 m/s. It accelerates with a constant upward acceleration

marta [7]

Answer:

Maximum height reached by the rocket, h = 202.62 meters

Explanation:

It is given that,

Initial speed of the model rocket, u = 56.5 m/s

Constant upward acceleration, $a=1.96\ m/s^2$

Distance traveled by the engine until it stops, d = 198.8 m

Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :

$v=\sqrt{u^2+2ad}$

$v=\sqrt{(56.5)^2+2\times 1.96\times 198.8}$

v = 63.02 m/s

At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.

Again using third equation of motion as :

$v^2-u^2=-2gh$

$h=\dfrac{v^2-u^2}{-2g}$

$h=\dfrac{u^2}{2g}$

$h=\dfrac{(63.02)^2}{2\times 9.8}$

h = 202.62 meters

So, the maximum height reached by the rocket is 202.62 meters. Hence, this is the required solution.

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2 years ago

A 50kg block is at rest on the ground.

yKpoI14uk [10]

Answer:

(b) 490 N downwards

Explanation:

(b) Formula : w= mg

m is the mass of the object

g is a gravitational constant (9.8)

w= mg

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