Answer:
lowers :)
Explanation:
What is the Tesla? How is it defined? Mention any other unit for the magnetic field.
Answer:
Tesla is name of a car
uu the way to go back and said that
To obtain maximum Electromotive force (EMF), you should connect the batteries in
Answer:
Series
Cells in Series connection.In series, cells are joined end to end so that the same current flows through each cell. In case if the cells are connected in series the emf of the battery connected to the sum of the emf of the individual cell,If E is the overall emf of the battery combined with n number cells and E1, E2,......Em is the EMFs of individual cell.
Then
E= E1+E2+...............+Em.
A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block. Round off your answer to zero decimal places.
Answer:
The answer is "512 J".
Explanation:
bullet mass [tex]m_1 = 10 g= 10^{-2} \ kg\\\\[/tex]
initial speed [tex]u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\[/tex]
block mass [tex]m_2 = 4\ Kg[/tex]
initial speed [tex]v_2 =-4.2 \frac{m}{s}[/tex]
final speed [tex]v_2= 0[/tex]
Let [tex]v_1[/tex] will be the bullet speed after collision:
throughout the consevation the linear moemuntum
[tex]\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\[/tex]
[tex]= 320 \frac{m}{s}[/tex]
The kinetic energy of the bullet in its emerges from the block
[tex]k=\frac{1}{2} m_1 v_1^2[/tex]
[tex]=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J[/tex]
A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the moment of inertia of the plate about point
The question is incomplete. The complete question is :
A plate of uniform areal density [tex]$\rho = 2 \ kg/m^2$[/tex] is bounded by the four curves:
[tex]$y = -x^2+4x-5m$[/tex]
[tex]$y = x^2+4x+6m$[/tex]
[tex]$x=1 \ m$[/tex]
[tex]$x=2 \ m$[/tex]
where x and y are in meters. Point [tex]$P$[/tex] has coordinates [tex]$P_x=1 \ m$[/tex] and [tex]$P_y=-2 \ m$[/tex]. What is the moment of inertia [tex]$I_P$[/tex] of the plate about the point [tex]$P$[/tex] ?
Solution :
Given :
[tex]$y = -x^2+4x-5$[/tex]
[tex]$y = x^2+4x+6$[/tex]
[tex]$x=1 $[/tex]
[tex]$x=2 $[/tex]
and [tex]$\rho = 2 \ kg/m^2$[/tex] , [tex]$P_x=1 \ $[/tex] , [tex]$P_y=-2 \ $[/tex].
So,
[tex]$dI = dmr^2$[/tex]
[tex]$dI = \rho \ dA \ r^2$[/tex] , [tex]$r=\sqrt{(x-1)^2+(y+2)^2}$[/tex]
[tex]$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$[/tex]
[tex]$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$[/tex]
[tex]$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$[/tex]
[tex]$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$[/tex]
[tex]$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$[/tex]
[tex]$I=\frac{32027}{21} \times 2$[/tex]
[tex]$= 3050.19 \ kg \ m^2$[/tex]
So the moment of inertia is [tex]$3050.19 \ kg \ m^2$[/tex].
Find the frequency and wavelength of the wave below, assuming it has a speed of 30 m/s
Which form of energy is responsible for the change of state here?
Answer:
d
Explanation:
A car goes 20 mph to 10 mph in five seconds find its acceleration
If a car travels 300 kilometers in 3 hours, its average speed is 100 km/hr.
O True
O False
Answer:
True
Explanation:
300/3=100km/hr
hope it helps. Plz mark me as brainliest.
Answer:
True
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