The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by

where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is

where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:

where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,

So the closest answer among the option is -5020 N.
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Answer: 0.85 meters (with and without sigfigs)
Explanation: To find the wavelength, you just have to switch around the equation for wave speed: v (wave speed) = λ (wavelength)*f (frequency) so λ (wavelength) = v (wave speed)/f (frequency). You don't have the wave speed but you can calculate it. Since wave speed is measured in meters/second or m/s, you just have to divide the amount of meters you were given by the amount of seconds. You will get 340 m/s. Next, you have to plug the values into the equation: λ (wavelength) = 340 m/s (wave speed)/400 Hz (frequency). The answer is 0.85 meters (seconds cancel) and has the correct number of significant figures.
Answer:
Δ
= 84 Ω,
= (40 ± 8) 10¹ Ω
Explanation:
The formula for parallel equivalent resistance is
1 /
= ∑ 1 / Ri
In our case we use a resistance of each
R₁ = 500 ± 50 Ω
R₂ = 2000 ± 5%
This percentage equals
0.05 = ΔR₂ / R₂
ΔR₂ = 0.05 R₂
ΔR₂ = 0.05 2000 = 100 Ω
We write the resistance
R₂ = 2000 ± 100 Ω
We apply the initial formula
1 /
= 1 / R₁ + 1 / R₂
1 /
= 1/500 + 1/2000 = 0.0025
= 400 Ω
Let's look for the error (uncertainly) of Re
= R₁R₂ / (R₁ + R₂)
R’= R₁ + R₂
= R₁R₂ / R’
Let's look for the uncertainty of this equation
Δ
/
= ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’
The uncertainty of a sum is
ΔR’= ΔR₁ + ΔR₂
We substitute the values
Δ
/ 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)
Δ
/ 400 = 0.1 + 0.05 + 0.06
Δ
= 0.21 400
Δ
= 84 Ω
Let's write the resistance value with the correct significant figures
= (40 ± 8) 10¹ Ω
Given:
mass is 3.1 kilograms
The acceleration due to gravity
is 9.8m/s2
Required:
Weight
Solution:
W = mg
W = (3.1 kilograms)( 9.8m/s2)
W = 30.38 Newtons
Newton’s second law is a=F/m this is what we will be using to solve this
However first you need to convert g to kg
100g= 0.1kg
0.05/0.1=0.5 m/s^2