Answer: 4.7m/s²
Explanation:
According to newton's first law,
Force = mass × acceleration
Since we are given more the one force, we will take the resultant of the two vectors.
Mass = 2.0kg
F1+F2 = (3i-8j)+(5i+3j)
Adding component wise, we have;
F1+F2 = 3i+5i-8j+3j
F1+F2 = 8i-5j
Resultant of the sum of the forces will be;
R² = (8i)²+(-5j)²
Since i.i = j.j = 1
R² = 8²+5²
R² = 64+25
R² = 89
R = √89
R = 9.4N
Our resultant force = 9.4N
Substituting in the formula
F = ma
9.4 = 2a
a = 9.4/2
a = 4.7m/s²
Therefore, magnitude of the acceleration of the particle is 4.7m/s²
Answer:
a) 4.31 m/s²
b) 215.5 m
Explanation:
a) According to Newton's first law of motion
The net force applied to particular mass produced acceleration, a, according to
F = ma
F = 140 N
m = 32.5 kg
a = ?
140 = 32.5 × a
a = 140/32.5 = 4.31 m/s²
b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s
u = initial velocity of the probe = 0 m/s (since it was initially at rest)
a = 4.31 m/s²
t = 10 s
s = distance travelled = ?
s = ut + at²/2
s = 0 + (4.31×10²)/2 = 215.5 m
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
45km/h * 0.5h= 22.5km
The car can travel 22.5km in 0.5 hours
Answer:
h~=371.26m
Explanation:
when an object falls we use the equations of accelerated motion. There is only one that gives distance.
Since we have no initial velocity (started from rest) we can get rid of the (ut) term
where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).
