A 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?

A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between

mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

5 0

3 years ago

Suzie Spacewalker hovers in space beside a rotating space station in outer space. Both she and the center of mass of the space s

Inga [223]

Answer:

is in the earths orbit

Explanation:

for Suzie to hover in space beside the rotating space station, she and the center of mass of the space station are at relative rest which happens when space station is in Earth orbit, hence she is in the earths orbit.

3 0

3 years ago

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

8 0

4 years ago

Hi guys i need truly help!

lutik1710 [3]

Answer:

1) 327

2) 3

3) 109

------------

now I'm not 100% sure #4-6 are correct, but if 4 is right then 5 and 6 are, too.

4) 1074

5) 1074 + 109t

6) 2382

Explanation:

7 0

3 years ago

The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click in

sergejj [24]

Answer:

The acceleration of the ball is $a_y = - 0.3672 \ m/s^2$

Explanation:

From the question we are told that

The maximum height the ball reachs is $H_{max} = 42.24 \ m$

The horizontal component of the initial velocity of the ball is $v_{ix} = 5.57 \ m/s$

The vertical component of the initial velocity of the ball is $v_{iy} = = 16.18 m/s$

The vertically motion of the ball can be mathematically represented as

$v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}$

Here the final velocity at the maximum height is zero so $v_{fy} = 0 \ m/s$

Making the acceleration $a_y$ the subject we have

$a_y = \frac{v_{iy} ^2}{2H_{max}}$

substituting values

$a_y = - \frac{5.57^2}{2* 42.24}$

$a_y = - 0.3672 \ m/s^2$

The negative sign shows that the direction of the acceleration is in the negative y-axis

6 0

3 years ago