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3 years ago

A 22.3 N crate is resting on the floor.

Physics

1 answer:

vitfil [10]3 years ago

8 0

weight of the crate is given as W = 22.3 N

now as per force balance condition

Net weight of the object is counter balanced by normal force on the object by the surface

so here we will have

$W = F_n$

$F_n = 22.3 N$

So the normal force acting on the object will be same as as the weight of the object which is 22.3 N

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MATHPHYS HELP

rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0

3 years ago

A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w

Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

R² = 2.4

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

3 0

3 years ago

Use the Pythagorean theorem to answer the following question. A ping-pong ball is shot straight north from a popgun at 4.0 m/s.

MaRussiya [10]

Resultant is 5 m/s using the Pythagorean theorem

4 0

3 years ago

Read 2 more answers

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

B) 1.0 × 10^5 V

Explanation:

Electric Potential Due To Point Charges

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.