All categories

2 years ago

6. A 73-kg woman stands on a scale in an elevator. The

Physics

1 answer:

olga nikolaevna [1]2 years ago

3 0

Answer:

Explanation:

n = m(g +a)

n= normal force (N)

m=mass (kg)

g=acceleration of gravity

a= acceleration of elevator

rearrange:

a= n/m - g

a= (810 N/73 kg) - 9.8 m/s ^2

a= 1.3 m/s ^2 up

and the acceleration is upwards bc her weight is less than the scale reading

You might be interested in

A 5592 N piano is to be pushed up a(n) 3.79 m frictionless plank that makes an angle of 30.1 ◦ with the horizontal. Calculate th

otez555 [7]

Answer:

10628.87 J

Explanation:

We are given that

Force applied =F=5592 N

$\theta=30.1^{\circ}$

Displacement=D=3.79 m

We have to find the work done in sliding the piano up the plank at a slow constant rate.

Work done= $F\times displacement$

The perpendicular component of force= $FSin\theta=5592sin(30.1)=2804.45N$

Work done = $Fsin\theta\times D=2804.45\times 3.79=10628.87 J$

Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J

8 0

2 years ago

Read 2 more answers

A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of Io in the coiled wire produces a magnet

Romashka-Z-Leto [24]

Answer:

vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Explanation:

The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation

B₀=μ₀nI₀

Where,

B₀ is the magnetic field in the middle of solenoid

n is the number of turns in the coil of solenoid

I₀ is the current flowing in the coil of solenoid

In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.

As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.

3 0

3 years ago

Read 2 more answers

Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200

artcher [175]

We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

6 0

3 years ago

Read 2 more answers

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

$\theta _2 = 13^o$

$\theta _1 =32.94^o$

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

2 years ago

Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha

Sedaia [141]

Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

The correct option is b

Explanation:

From the question we are told that

The charge carried by A is q C

The charge carried by B is 0 C

The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by A and B is mathematically evaluated as

$\frac{ 0 + q}{2} = \frac{q}{2}$

When C and B are brought close and then separated the charge carried by C and B is mathematically evaluated as

$\frac{0 + \frac{q}{2} }{2} = \frac{q}{4}$

When C and A are brought close and then separated the charge carried by C and A is mathematically evaluated as

$\frac{\frac{q}{4} + \frac{q}{2} }{2} = \frac{3q}{8}$

Looking at these calculation we can see that the charge carried by B is

$\frac{q}{4} C$

8 0

2 years ago