Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)
Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N
-F
towards the sister (-F) (greater force applied)
Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Answer:
Magnetic force, F = 0.24 N
Explanation:
It is given that,
Current flowing in the wire, I = 4 A
Length of the wire, L = 20 cm = 0.2 m
Magnetic field, B = 0.6 T
Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :


F = 0.24 N
So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.
According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

Here,
is the velocity of water through wide ends of cylindrical pipe and
is the velocity of water through narrow ends of cylindrical pipe.
Given, 
Now from equation continuity,
.
Here,
and
are cross- sectional areas of wide and narrow ends of cylindrical pipe.
As pipe is circular, so
.
At the second point, the diameter is halved, which means the radius is also halved. Therefore,


Substituting these values with the density of water is
in pressure difference formula we get.

Answer:
a) 0.0288 grams
b) 
Explanation:
Given that:
A typical human body contains about 3.0 grams of Potassium per kilogram of body mass
The abundance for the three isotopes are:
Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.
a)
Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.
However, the amount of potassium that is present in such person is :
0.012% × 240 grams
= 0.012/100 × 240 grams
= 0.0288 grams
b)
the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:
First the Dose in (Gy) = 
= 
= 
Effective dose (Sv) = RBE × Dose in Gy
Effective dose (Sv) = 
Effective dose (Sv) = 