Answer:
The distance traveled by the balloon is 10.77 m
Explanation:
velocity of the ball, 
= 2 m/s south
velocity of the air, 
= 5 m/s west
To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.
| 2m/s
|
|
↓
5m/s ←------------------
the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.
R² = 5² + 2²
R² = 29
R = √29
R = 5.385 m/s
The distance traveled by the balloon is calculated as;
d = R x t
where;
t is time of the motion = 2 seconds
d = 5.385 x 2
d = 10.77 m
Therefore, the distance traveled by the balloon is 10.77 m.
Answer:
v = 1.2 m/s
Explanation:
The wavelength of the waves is given as the horizontal distance between the crests:
λ = wavelength = 5.5 m
Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:
T = Time Period = 2(2.3 s) = 4.6 s
Now, the speed of the wave is given as:
where,
v= speed of wave = ?
f = frequency of wave = 
Therefore,
<u>v = 1.2 m/s</u>
Explanation:
it is energy that flows of electric charge it has the ability to do work or apply force to move an object.
The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
Answer:
k1 + k2
Explanation:
Spring 1 has spring constant k1
Spring 2 has spring constant k2
After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.
x1 = x2
Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are
k1 = F1/x -> F1 =k1*x
k2 = F2/x -> F2 =k2*x
While F = F1 + F2
Substitute equation of F1 and F2 into the equation of sum of forces
F = F1 + F2
F = k1*x + k2*x
= x(k1 + k2)
Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)
Considering the general equation of spring forces (Hooke's Law) F = kx,
The effective spring constant for the system is k1 + k2
