The graph shows the federal budget from 1980 to 2010. In which period did the federal budget show the greatest deficit?

The game was played by swinging the big mallet down hard enough to cause the bell to ring. If it took a 44 newton force to ring

nata0808 [166]

I think C. 33 newtons

3 0

3 years ago

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m

erastovalidia [21]

Answer:

a) W = 46.8 J and b) v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

W = ΔK = $k_{f}$ -K₀

a) work is the scalar product of force by distance

W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

W = F d cos θ

W = 39.0 1.20 cos 0

W = 46.8 J

b) zero initial kinetic language because the package is stopped

W - $W_{fr}$ = $k_{f}$ -K₀

W - fr d= ½ m v² - 0

W - μ N d = ½ m v

on the horizontal surface using Newton's second law

N-W = 0

N = W = mg

W - μ mg d = ½ m v

v² = (W -μ mg d) 2/m

v = √(W -μ mg d) 2/m

v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3 ]

v = √(31.63 2/4.3)

v = 3.84 m/s

8 0

3 years ago

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

3 years ago

How is the magnetic force on a particle moving in a magnetic field different from gravitational and electric forces.

harina [27]

Answer:

The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field, direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.

Explanation:

The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field and direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.

The magnetic force is given by the charge times the vector product of velocity and magnetic field. While gravitational force is given by the square of the particle mass divided by the square its distance of separation. Also electric forces is given by the square of the charge magnitude divided by the square its distance separation.

4 0

3 years ago

What surfaces are good absorbers of infrared radiation?

WINSTONCH [101]

Black surfaces is you'r answer. or dark mater. ya those are you'r answer.

8 0

3 years ago

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