3 years ago

Which technology contains at least one permanent magnet?

Physics

2 answers:

stich3 [128]3 years ago

6 0

I’m pretty sure it’s, B. An Electric Motor. I apologize in advance if it’s incorrect.

Xelga [282]3 years ago

4 0

Answer: B

Explanation: took it on ape x I know the other guy said it this is just for reassurance

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How long must a 400w electrical engine work in order to produce 300kj of work

vladimir1956 [14]

Answer:

300000 J / 400 J/s = 750 s = 12.5 minutes

Explanation:

8 0

2 years ago

The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please

beks73 [17]

I'm pretty sure it's D. The stars don't influence the moon's phases.

6 0

3 years ago

the imagine above shows to opposite forces acting on a rolling cart, what can we say is true about affect of the forces on the c

cluponka [151]

Answer:

Here is my answer...

Explanation:

The cart will connect with the opposite force, and then the cart will come to a shuddering stop before moving in the direction of the oposite force.

Hope I helped! :)

7 0

2 years ago

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

3 years ago

Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat

svetlana [45]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

6 0

2 years ago