Answer:
Average force = 3.5 kN
Explanation:
Given:
Mass of Jennifer (m) = 50 kg
Initial velocity = 35 m/s
Time taken to stop body = 0.5 s
Find:
Average force
Computation:
v = u + at
0 = 35 + a(0.5)
Acceleration (a) = - 70 m/s² = 70 m/s²
Average force = ma
Average force = (50(70)
Average force = 3500 N
Average force = 3.5 kN
Answer:
- 3 cm
Explanation:
From the mirror formula;
1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.
1/-4.5 = 1/9 + 1/v
1/v = -1/4.5 - 1/9
= -1/3
Therefore;
v = -3 cm
Hence;
Image distance is - 3cm
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
