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ElenaW [278]
3 years ago
11

Which of the following unit is the unit of resistance? * 1 point A. A B. Ω C. V D. R

Physics
1 answer:
tekilochka [14]3 years ago
5 0

Answer:

its b

Explanation:

ohm is the unit of resistance

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A new landowner has a triangular piece of flat land she wishes to fence. Starting at the first corner, she measures the first si
Lera25 [3.4K]

Answer:

Length of third side = 3.97m

Explanation:

First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.

From A-B we have distance = 5.5m = Say it c

From B-C we have distance = 4.3m = Say it a

From A-C we have distance = ? = Say it b which we have to find out.

Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)

For our case it is:

b² = a² + c² - 2acCos(B)         -  Say it equation 1

From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.

There values are:

a = 4.3m;  c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees

Now by putting the respectve values in equation 1 we have:

b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)

b² = 18.49 + 30.25 - 32.95

b² = 15.79

b  = √15.79

b = 3.97m

Thus the length of third side is 3.97m.

PS: The picture of triangle is being attached for yours understanding.

6 0
3 years ago
A water tank holds water to the depth of the 80cm what is the pressure of the water of the tank​
Mashutka [201]

Answer:

7976 Pascals significant figure= 7.9*10^3

Explanation:

formula of hpg = height*density*gravitational energy

.80*10*997=7976 pascals

8 0
2 years ago
The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25
Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0
3 years ago
A car of weight 300N moved through a distance of 10m when pushed by 3 students
Sati [7]

The work done by the three students is 3,000 J.

The energy transferred in the process is 3,000 J.

<h3>What is work done?</h3>
  • Work done is the product of force and distance moved by the object.

W = Fd

The work done by the three students is calculated as follows;

W = 300 x 10

W = 3,000 J

<h3>What is energy transfer?</h3>
  • This is means by which energy is converted from one form to another.

The energy transferred in the process is determined by work energy theorem.

E = W

E = 3,000 J

Learn more about work-energy theorem here: brainly.com/question/22236101

3 0
2 years ago
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
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