The net force of the cart when it is pushed to the right with a force of 15N.
<u>Explanation:</u>
To find the force of net, which is calculated by the formula.
The Net Force= Addition of the force applied on the respective direction.
The Net Force here is given by
The Net Force = 15-20 (A force towards the right and a force towards left, two opposite so subtraction).
Hence
Thus the Net Force = -5(The force towards left, so it gets a negative value).
Answer:
C) 20 m/s
Explanation:
Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.
The velocity of a moving wave is
v = λf ............................ Equation 1
Where v = speed of the wave, λ = wave length, f = frequency of the wave.
Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters
Substituting these values into equation 1
v = 2×10
v = 20 m/s.
Thus the speed of the wave = 20 m/s
The right option is C) 20 m/s
The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L