Answer:
(a) -202 m/s²
(b) 198 m
Explanation:
Given data
- Initial speed (v₀): 283 m/s
- Final speed (vf): 0 (rest)
(a) The acceleration (a) is the change in the speed over the time elapsed.
a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²
(b) We can find the distance traveled (d) using the following kinematic expression.
y = v₀ × t + 1/2 × a × t²
y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²
y = 198 m
Answer:
(a) FN = m (g - 
)
(b) vmin = 17.146 m/s
Explanation:
The radius of the arc is
r = 30m
The normal force acting on the car form the highest point is
FN = m (g - )
If the normal force become 0 we have
m (g - ) = 0
or
g - = 0
This way, when FN = 0, then v = vmin, so
g - 
= 0
vmin = ![\sqrt[.]{g*r}](https://tex.z-dn.net/?f=%5Csqrt%5B.%5D%7Bg%2Ar%7D)
= ![\sqrt[.]{9.8 m/s^{2} * 30m } = 17.146 m/s](https://tex.z-dn.net/?f=%5Csqrt%5B.%5D%7B9.8%20m%2Fs%5E%7B2%7D%20%2A%2030m%20%7D%20%3D%2017.146%20m%2Fs)
If a cruise ship is having troubles with buoyancy, then spread the weight of the ship over a greater volume.
Answer: Option D
<u>Explanation:
</u>
Buoyancy is the upward thrusting phenomenon of water acting on any object immersed partially or fully in water body. Hence, it creates the buoyant forces that is inversely proportionate to the immersing body's density. If the immersing body's density is higher than the density of the immersing medium then the body will get completely immersed in the water.
Similarly, in case of less, the buoyant forces act on the body will prevent it from complete immersion and allow it to float on water. Mostly cruise ships and other navy vessels use this phenomenon to keep on floating on surface of water.
In the present condition, the solution for buoyancy problem faced by a cruise ship can be solved by decreasing the density of the ship. And the ship's density can be decreased by increasing the ship's volume or by spreading the ship's weight over a greater volume.
The image distance when a boy holds a toy soldier in front of a concave mirror, with a focal length of 0.45 m. is -0.56 m.
<h3>What is image distance?</h3>
This is the distance between the image formed and the focus when an object is placed in front of a plane mirror.
To calculate the image distance, we use the formula below.
Formula:
- 1/f = 1/u+1/v........... Equation 1
Where:
- f = Focal length of the mirror
- v = Image distance
- u = object distance
From the question,
Given:
Substitute these values into equation 1 and solve for the image distance
- 1/0.45 = 1/0.25 + 1/v
- 2.22 = 4+1/v
- 1/v = 2.22-4
- 1/v = -1.78
- v = 1/(-1.78)
- v = -0.56 m
Hence, The image distance is -0.56 m.
Learn more about image distance here: brainly.com/question/17273444
Answer:
76.1N
Explanation:
Given parameters:
Mass of the ball = 7.77kg
Unknow:
Weight of balloon = ?
Solution:
Weight is the vertical force applied on a body.
Weight = mass x acceleration due gravity
So;
Weight = mass x acceleration due to gravity
So;
Weight = 7.77 x 9.8 = 76.1N
