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3 years ago

What is the y component of a vector defined as 12.2m at 81.5°?

Physics

1 answer:

sergejj [24]3 years ago

3 0

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

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3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left

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Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v

where v is the velocity of the 2.22 kg block after collision. Solve for v :

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A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic frict

Firdavs [7]

Answer:

t = 23.92 s

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction, downward

FN : Normal force : perpendicular to the floor, upward

fk : Kinetic friction force: parallel to the floor and opposite to the movement

F = 86.4 N , in the direction of the motion

Calculated of the W

W= m*g

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Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - W = 0

FN = W

FN = 164.64 N

Calculated of the fk

fk = μk*FN

fk = 0.426* 164.64 N

fk = 70.13 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F - fk = m*a

86.4 -70.13 = (16.8)*(-a)

16.26 = (16.8)*(-a)

a = -(16.26 )/ (16.8)

a = - 0.97 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula :

vf = v₀ + a*t Formula (2)

Where:

t: time interval (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 23.2 m/s

vf = 0

a = -0.97 m/s²

Time it takes for the block to stop

We replace data in the formula (2) to calculate the time

vf= v₀+a*t

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t = 23.2 / (0.97)

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Answer:

297.8 m

Explanation:

We are given that

Acceleration= $a=1.65m/s^2$

Time,t=19 s

We have to find the distance covered by the train when it reach its top speed if starting from rest.

Initial speed=u=0

$s=ut+\frac{1}{2}at^2$

Using the formula

$s=0(19)+\frac{1}{2}\times 1.65(19)^2$

$s=297.8 m$

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