Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
Answer:
So airplane will be 1324.9453 m apart after 2.9 hour
Explanation:
So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 = 
Distance traveled by first plane = 730×2.9 = 2117 m
And distance traveled by second plane = 590×2.9 = 1711 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.
Using the law of cosine,
representing the distance between the planes, we see that:

d = 1324.9453 m
Answer:
The resistance of the inductor at resonance is 258.76 ohms.
Explanation:
Given;
resistance of the resistor, R = 305 ohm
capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F
inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H
At resonance the inductive reactance is equal to capacitive reactance.

Where;
F₀ is the resonance frequency

The inductive reactance is given by;

Therefore, the resistance of the inductor at resonance is 258.76 ohms.
Explanation:
As a tornado or other storm system passes over a building, low pressure can tug a roof upward. When those forces surpass the force exerted by the weight of the roof, the structure flies up and is swept away by wind currents..
Answer:
a) P = 44850 N
b) 
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:

on substituting the values, we get

or
Load, P = 44850 N
Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>
b)The deformation (
) due to an axial load is given as:

on substituting the values, we get

or
