Answer:
13 m/s
Explanation:
I assume we are ignoring friction.
The boy's PE will all be converted to KE at the bottom of the hill.
to find PE = mgh we need to know h
h = 50 sin 10 = 8.68 meters
then: PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j
KE = 1/2 m v^2 = <u>1703 .49</u>
v = 13 m/s
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:
The average net force on the truck is 375 Newtons.
Explanation:
Using Newton's 3rd equation of motion, we have :
×a×s
where, v = final velocity = 25 m/s
u = initial velocity = 20 m/s
a = acceleration
s = distance traveled = 300 m
Using these values in the above equation, we get acceleration = 0.375 m/
Using Newton's second law, we have:
F=m×a
where m = mass = 1000 kg
a= acceleration = 0.375 m/
Putting values we have F=375 N
Answer:
a = -0.33 m/s² k^
Direction: negative
Explanation:
From Newton's law of motion, we know that;
F = ma
Now, from magnetic fields, we know that;. F = qVB
Thus;
ma = qVB
Where;
m is mass
a is acceleration
q is charge
V is velocity
B is magnetic field
We are given;
m = 1.81 × 10^(−3) kg
q = 1.22 × 10 ^(−8) C
V = (3.00 × 10⁴ m/s) ȷ^.
B = (1.63T) ı^ + (0.980T) ȷ^
Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;
a = qVB/m
a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))
From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^
Thus;
a = -0.33 m/s² k^
