Answer:
the answer is D
Explanation:
Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.
Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass 
= 30 kg
mass 
= 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =![[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]](https://tex.z-dn.net/?f=%5B%2050%20%20-%2030%20%2A%20%5Cfrac%7B1%7D%7B2%7D%20-%200.1%20%2A30%2A%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D)
g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
The answer would be 20000
The answer in standard form/scientific notation would be 2 x 10^4
(The exponent is 4 because that's how many digits after the 2 there is)
