Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So 
The velocity is given by:
V = √(Vx²+Vy²)
V = velocity, Vx = horizontal velocity, Vy = vertical velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for V:
V = √(6²+12²)
V = 13.42m/s
Now find the direction:
θ = tan⁻¹(Vy/Vx)
θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity
Given values:
Vx = 6m/s, Vy = 12m/s
Plug in and solve for θ:
θ = tan⁻¹(12/6)
θ = 63.4°
The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.
Answer:
(a) T = 0.015 N
(b) M = 1.53 x 10⁻³ kg = 1.53 g
Explanation:
(a) T = 0.015 N
First, we will find the speed of waves:

where,
v = speed of wave = ?
f = frequency = 120 Hz
λ = wavelength = 6 cm = 0.06 m
Therefore,
v = (120 Hz)(0.06 m)
v = 7.2 m/s
Now, we will find the linear mass density of the coil:

where,
μ = linear mass density = ?
m = mass = 1.45 g = 1.45 x 10⁻³ kg
l = length = 5 m
Thereforre,

Now, for the tension we use the formula:

<u>T = 0.015 N</u>
<u></u>
(b)
The mass to be hung is:

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>
Answer:
A. 243 N
Explanation:
Friction is the force that opposes the relative motion between systems that are in contact.
This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.
First find the normal force which is the force that would point directly upwards to support weight of the block.
Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.
Given m=40 kg and g=9.80 m/s²
N force=40×9.80 =392N
Then find the force of friction which is given by the formula;
<em>F=μN where μ is friction coefficient for the oak chest and N is the normal force on the chest</em>
Given <em>μ</em>=0.620 and N force = 392 N then it will be;
F=0.620× 392 =243.04 N
Answer : 243 N