Ask question

Ask question

All categories

3 years ago

7

You are standing in an elevator that is accelerating upward at 2 m/s2. What is the normal force acting on you by the elevator if

your mass is 70 kg?

1 answer:

Delvig [45]3 years ago

6 0

Answer:

784 N.

Explanation:

The expression for the Normal reaction acting on a body inside an elevator that is moving upward is given as

R = m(g+a)................. Equation 1

Where R = normal reaction, g = acceleration due to gravity, a = acceleration, m = mass .

Given: m = 70 kg, a = 2 m/s²

Constant: g = 9.8 m/s²

Substitute into equation 1

R = 70(2+9.2)

R = 70(11.2)

R = 784 N.

Hence the normal force = 784 N.

You might be interested in

What is 34 + (5) × (1.2465) written with the correct number of significant figures?

bonufazy [111]

Answer: 40

Explanation:

= 34 + 5 * 1.2465

= ‭40.2325‬

= 40

The number of significant figures in the answer should be the same as the number with the least number of significant figures that any of the digits in the equation have.

32 has 2 significant figures so the answer has to be 2 significant figures which is 40.

7 0

4 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago

According to the picture, which is the least dense?

alexdok [17]

Answer:

a the chess peice

Explanation:

my head

4 0

3 years ago

Read 2 more answers

Which of the following is true about all of the outer planets?A.Much of the material in these planets is solid.B.The surface of

Lisa [10]

The awnsers is D all the planet's in the solar system rotate the same direction or else they would bump into each other

4 0

3 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

8 0

3 years ago