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zysi [14]
3 years ago
10

Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m. Which statement a

ccurately describes something that Yolanda can do as a part of her study?a. She can use destructive interference to generate a wave with an amplitude of 3.1 m. b. She can use constructive interference to generate a wave with an amplitude of 1.5 m. c. She can use destructive interference to generate a wave with an amplitude of 2.8 m. d. She can use constructive interference to generate a wave with an amplitude of 3.5 m.
Physics
1 answer:
Anna71 [15]3 years ago
8 0

Answer:

a) Not Accurate

b) Not Accurate

c) Accurate

d) Accurate

Explanation:

Part a

Not Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m

Part b

Not Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m

Part c

Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m by varying the phase difference between two waves she can achieve the desired results.

Part d  

Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m by varying the phase difference between two waves she can achieve the desired results.

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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
if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav
allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, R_{max} = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, R_{max}, of a projectile motion is found by using the following formula

R_{max} = \dfrac{u^2}{g}

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, g_{Moon} = 1/16 × g

∴ g_{Moon} = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

R_{max \ Moon} = \dfrac{u^2}{g_{Moon}}   = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is R_{max \ Moon} ≈ 48 m

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