m = 5 kg
a = 2 m/s²
to find the force that accelerates the 4 kg object @ 2 m/s²
F = ma = 5 kg x 2 m/s² = 10 N
To find what acceleration 10 N would give a 20 kg object
a = F/m = 10 N/20 kg = 0.5 m/s
Answer:
Explanation:
There are 3 main forces at work here, gravity, normal and friction. The gravity pulls the car straight down and is what keeps the car on the ground. Normal force is straight up from the points where the car is touching, so since the wheels are the only parts of the car touching the street, this is where all the normal force is. Friction force opposes any and all motion, the car wants to slide down the hill and would slide down the hill if there was no friction, so the friction force is in the opposite direction of the cars intended motion.
Answer:
F = - 3.53 10⁵ N
Explanation:
This problem must be solved using the relationship between momentum and the amount of movement.
I = F t = Δp
To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio
v = d / t
t = d / v
Reduce SI system
m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg
d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m
Let's calculate
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
With this value we use the momentum and momentum relationship
F t = m v - m v₀
As the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵
F = - 3.53 10⁵ N
The negative sign indicates that the force is exerted against the bullet
Answer:
0.0667 m
Explanation:
λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m
D = screen distance = 2.5 m
d = slit width = 15 x 10⁻⁶ m
n = order = 1
θ = angle = ?
Using the equation
d Sinθ = n λ
(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)
Sinθ = 26.67 x 10⁻³
y = position of first minimum
Using the equation for small angles
tanθ = Sinθ = y/D
26.67 x 10⁻³ = y/2.5
y = 0.0667 m
Answer:
Once used as a energy source they cannot be charged/used again.
Explanation:
