Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
Answer:
12.6332454263 m/s
Explanation:
m = Mass of car
v = Velocity of the car
= Coefficient of static friction = 0.638
g = Acceleration due to gravity = 9.81 m/s²
r = Radius of turn = 25.5 m
When the car is on the verge of sliding we have the force equation

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s
Answer:

Explanation:
From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

Since we know gravity and its hight

Answer:
The magnitude of magnetic field at given point =
×
T
Explanation:
Given :
Current passing through both wires = 5.0 A
Separation between both wires = 8.0 cm
We have to find magnetic field at a point which is 5 cm from any of wires.
From biot savert law,
We know the magnetic field due to long parallel wires.
⇒ 
Where
magnetic field due to long wires,
,
perpendicular distance from wire to given point
From any one wire
5 cm,
3 cm
so we write,
∴ 

![B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%204%5Cpi%20%5Ctimes10%5E%7B-7%7D%20%5Ctimes5%7D%7B2%5Cpi%20%7D%20%5B%5Cfrac%7B1%7D%7B0.03%7D%20%2B%20%5Cfrac%7B1%7D%7B0.05%7D%20%5D)

Therefore, the magnitude of magnetic field at given point = 
Answer:

Explanation:
given,
frequency of tuba.f = 64 Hz
Speed of train approaching, v = 8.50 m/s
beat frequency = ?
using Doppler's effect formula

v_s is the velocity of the source
v is the speed of sound, v = 340 m/s
now,

f' = 65.64 Hz
now, beat frequency is equal to



hence, beat frequency is equal to 1.64 Hz