Answer:
<h3>The answer is 15 N</h3>
Explanation:
The force acting on an object can be found by using the formula
<h3>Force = mass × acceleration</h3>
From the question
mass = 50 g = 0.05 kg
acceleration = 300 m/s²
We have
force = 0.05 × 300
We have the final answer as
<h3>15 N</h3>
Hope this helps you
The car's rate of acceleration : a = 2.04 m/s²
<h3>Further explanation</h3>
Given
speed = 110 km/hr
time = 15 s
Required
The acceleration
Solution
110 km/hr⇒30.56 m/s
Acceleration is the change in velocity over time
a = Δv : Δt
Input the value :
a = 30.56 m/s : 15 s
a = 2.04 m/s²
Answer:
14.2 m
Explanation:
Using conservation of energy:
PE at top = KE at bottom
mgh = ½ mv²
h = v² / (2g)
h = (16.7 m/s)² / (2 × 9.8 m/s²)
h = 14.2 m
Using kinematics:
Given:
v₀ = 16.7 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 14.2 m
Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
Explanation:
Stopping distance, s = 3 ft = 0.914 m
final velocity, v = 0
a = g/2 = 4.9 m/s²
Use third equation of motion:
substitute the values to find the speed of train:
<h2>Answer</h2>
option D)
2.4 seconds
<h2>Explanation</h2>
Given in the question,
mass of car = 1200kg
speed of car = 19m/s
Force due to direction of travel
F = ma
= 12000(a)
Force to due frictional force in reverse direction
-F = mg(friction coefficient)
= -12000(9.81)(0.8)
<h2>
-mg(friction coefficient) = ma </h2>
(cancelling mass from both side of equation)
g(0.8) = a
(9.81)(0.8) = a
a = 7.848 m/s²
<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>
where vf = final velocity
vo = initial velocity
a = acceleration
t = time
0 - 19 = 7.8(t)
t = 19/7.8
= 2.436 s
≈ 2.4s
