Answer: 9 m
Explanation:
Work is said to be done when an unbalanced force causes displacement of the body.
Force is the product of mass (m) and acceleration (a).
Work = Force × Displacement
⇒W = F.s = ma.s
It is given that mass of the dresser is, m = 250 kg
work done, W = 126 J
Force acting on the dresser, F = 14 N
we need to find displacement, s
⇒126 J = 14 N × s
⇒ s = 126 J/ 14 N = 9 m
Hence, Laurie is able to move the dresser to about 9 m.
Answer:
p = 20 kg•m/s
KE = 100 J
Explanation:
In an elastic collision of identical masses, the two masses will exchange momentums. Therefore Block 1 initially moving at 10 m/s will be moving at 2 m/s, and Block 2 will go from 2 m/s to 10 m/s
momentum = mv = 2(10) = 20 kg•m/s
KE = ½mv² = ½(2)10² = 100 J
Unfortunately, your answer selection does not have this answer as an option.
Answer:
The second one is a function. {(-8, -2), (7, -2),(-9,2), (0,0)
Explanation:
Its because the y-value is repeated twice.
Hope it helps.
Answer:
A. 59.4
Explanation:
The refractive index of the glass, n₁ = 1.50
The angle of incidence of the light, θ₁ = 35°
The refractive index of air, n₂ = 1.0
Snell's law states that n₁·sin(θ₁) = n₂·sin(θ₂)
Where;
θ₂ = The angle of refraction of the light, which is the angle the light will have when it passes from the glass into the air
Therefore;
θ₂ = arcsin(n₁·sin(θ₁)/n₂)
Plugging in the values of n₁, n₂ and θ₁ gives;
θ₂ = arcsin(1.50 × sin(35°)/1.0) ≈ 59.357551° ≈ 59.4°
The angle the light will have when it passes from the glass into the air, θ₂ ≈ 59.4°.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
