Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
d. less than 20m/s
Explanation:
To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is
[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s
As 14.14m/s is less than 20m/s. d is the correct selection for this question.
Answer:
60words/minute
Explanation:
If Sunitha can type 1800 words in half an hour, this can be expressed as;
1800 words = 30 minutes
To get her typing speed per minute, we will use the formula
Speed = Number of words/Time used
Typing speed = 1800/30
Typing speed = 60words/minute
Hence her typing speed in words per minute is 60words/minute
Answer:
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
Answer:
(a) 29 cm
(b) 43.5 cm
Explanation:
(a) when loop A is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 70 cm mark
-2 N at x
Taking the sum of the torques about B:
∑τ = Iα
(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0
18 Ncm − 2 N (x − 70 cm) = 0
2 N (x − 70 cm) = 18 Ncm
x − 70 cm = 9 cm
x = 79 cm
The distance from the center is |50 cm − 79 cm| = 29 cm.
(b) when loop B is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 20 cm mark
-2 N at x
Taking the sum of the torques about A:
∑τ = Iα
(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0
-27 Ncm − 2 N (x − 20 cm) = 0
2 N (x − 20 cm) = -27 Ncm
x − 20 cm = -13.5 cm
x = 6.5 cm
The distance from the center is |50 cm − 6.5 cm| = 43.5 cm
