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2 years ago

A forward honzontal force of 50 N is applied to a crate. A second horizontal force o

Physics

1 answer:

True [87]2 years ago

4 0

Answer:

Explanation:

magnitude: 180-50=130N

Direction: in the direction same as the second horizontal force

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Two point charge

timama [110]

Answer:

Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?

Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?

4 0

1 year ago

A car travelling at a constant speed of 70km/h passes a stationary police car. The police car immediately goes on the chase acce

Virty [35]

Answer:

18.24 seconds

Explanation:

First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.

You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

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2 years ago

You accidentally drop an eraser out of the window of an apartment 15 m above the ground

docker41 [41]

Answer:

hello, yes or nou sorry jaja

3 0

2 years ago

If the period of a spring is 5 seconds what is the frequency

Scilla [17]

Answer:

C. 0.2 Hertz

Explanation:

The frequency of a spring is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

For the spring in this problem,

T = 5 s

therefore, the frequency is

$f=\frac{1}{5 s}=0.2 Hz$

7 0

2 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago