Answer:
the correct answer is E
A graph of the cart's maximum speed squared as a function of x^3
Explanation:
For this exercise let's use Newton's second law
F = m a
force has the form
F = k x²
and acceleration is related to velocity
a = dv / dt
Let's use the chain rule or L'Hospital
a = dv /dx dx/dt
a = dv /dx v
let's substitute
k x² = m v dv / dx
k /m x² dx = v dv
we integrate
k /m x³ /3 = v² / 2
v² = (2k /3m) x³
This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable
y = v²
x´ = x³
y = (2k/3m) x´
if we graph y vs x 'we have a linear graph whose slope is
m = 2k / 3m
By reviewing the different answers, the correct answer is E
Answer:
The inducerd emf is 1.08 V
Solution:
As per the question:
Altitude of the satellite, H = 400 km
Length of the antenna, l = 1.76 m
Magnetic field, B = 
Now,
When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:
Here, velocity v is perpendicular to the rod
Thus
e = lvB (1)
For the orbital velocity of the satellite at an altitude, H:
where
G = Gravitational constant
= mass of earth
= radius of earth
Using this value value in eqn (1):
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Answer:
The net force = 0
Explanation:
The given information includes;
The mass of the crate = 250 kg
The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)
In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.
The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate
The weight of the crate, 
↓ = 250 kg × 9.81 m/s² = 2,452.5 N
The force the helicopter should provide to just lift the crate, 
↑ = The weight of the crate = 2,452.5 N
The net force, 
= ↑ - ↓ = 2,452.5 N - 2,452.5 N = 0
The net force = 0.
Kinetic energy =0.5*mas*velocity^2
Joules =lg*m^2/s^2
1 miles= 1608.34 meters
1 hour= 3600 Sec
1 ounce =28.35g =0.02836 kg
What is a the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher at 96.0 mi/h?
Answer: KE=0.5m*v^2
=0.5*(5.12 o *0.02835 kg/1 ounce)* (95 miles/h*1609.34m/1 miles* 1hr/3600s^)2
131kg*m^2/s^2= 131 joules
By what factor with the kinetic energy change if the speed of the baseball is decreased to 55.0 mi/h?
Answer: KE=0.5*m*v^2
=0.5*(5.13 o*0.02835kg/1 ounce)*(55 miles/ h*1609.34m/1 mile*1 hr/3600s)^2
=44.0kg*m^2s^2=44.0 joules
131/44= 2.98, so decreased by a factor of approximately 3
