Formula for distance is d=st
so for speed is s=d/t
48 km per hour
Answer:
Vf= 3.435 m/s
Explanation:
Given:
Initial velocity Vi =0 m/s (starting from Rest position)
θ = 37⁰
Distance S = 1 m
To find: Final Velocity Vf=?
fist we have to find the down slope net acceleration a = g sin θ
a= 9.81 sin 37⁰ = 5.9 m/s²
By 3rd equation of motion
2 a S= Vf² - Vi²
Vf = Square root ( 2 × 5.9 m/s² × 1 + 0 m/s)
Vf = Square root (11.8)
Vf= 3.435 m/s
Answer:
2.89 hours
Explanation:
given :
Vo = 72 km/h
Vt = 90 km/h
S = 234 km
find : the time taken (t) = ?
solution :
2.a.s = Vt² - Vo²
2.a.(234) = 90²- 72²
468.a = 8100 - 5184
= 2916
a = 2916/468 = 6.23 km/h²
so,
t = (Vt-Vo) /a
= (90-72)/ 6.23
= 18/ 6.23
= 2.89 hours
Answer:
t₁ > t₂
Explanation:
A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁
Solution:
Newton's law of motion is given by:
s = ut + (1/2)gt²;
where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.
u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.
hence:
When stationary:

Hence t₂ < t₁, this means that t₁ > t₂.
Answer:
Distance: -30.0 cm; image is virtual, upright, enlarged
Explanation:
We can find the distance of the image using the lens equation:

where:
f = 15.0 cm is the focal length of the lens (positive for a converging lens)
p = 10.0 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q,

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).
The magnification can be found as

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.
Also, the fact that the magnification is positive tells us that the image is upright.