Anyone know the answer ?

A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15

wel

Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg : mass of the block

θ =37.5° :angle θ of the ramp with respect to the horizontal direction

μk= 0.460 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

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 on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W= 4.25 kg* 9.8 m/s² = 41.65 N

x-y weight components

Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

Wy= Wcos θ =41.65*cos 37.5° =33.04 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk* N= 0.460*33.04

f = 15.2 N

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

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

-Wx-f = m*a

-25.35-15.2 = (4.25)*a

-40.55 = (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:

d:displacement (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

2*(9.54)*d = (15)²

(19.08)*d = 225

d = 225 / (19.08)

d = 11.79 m

3 0

3 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

3 years ago

Please help (will mark brainliest)

serg [7]

Answer:

if im not mistaken i think its d let me know if correct plz

7 0

3 years ago

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Bob has been Springfield's top meteorologist for ten years. He received data from satellite imaging and computer models that a h

djyliett [7]

The correct letter of the answer would be letter b, Bob is using his personal knowledge of the local area to make his forecast. It does not look like he's ignoring the computer for he's been the top meteorologist for ten years, meaning to say, he would use his knowledge about the local area and tell them his knowledge about the weather.

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3 years ago

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A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha

kifflom [539]

Answer:

$x_{distance}=10.07km$