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3 years ago

The sum of the interior angles of a quadrilateral is _______. 180° 90° 270° 360°

Physics

2 answers:

nlexa [21]3 years ago

7 0

Year9 formula: 180 (n-2)
Answer: 360

olya-2409 [2.1K]3 years ago

4 0

Answer:

360°

Explanation:

Quadrilateral are polygons that have four sides. The sum of internal angle of a quadrilateral is 360°. The quadrilateral are closed shape with four sides. Examples of quadrilaterals are trapezium, rectangle, square, Rhombus and parallelograms.

Every quadrilaterals can be divided into two triangle. The sum of angle in a triangle is 180°. Adding the two triangle angle together , we will get the sum as 360°.

Sum of the interior angle of a polygon = 180(n-2) . In the case of quadrilateral, it has 4 sides. n represents the number of sides of the polygon .

interior angle of a quadrilateral = 180(4-2) = 360°

The image illustrate how the interior angle of quadrilaterals can be sum.

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Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co

balu736 [363]

The combined momentum of the passengers is 5000 kgm/s.

<h3>Combined momentum of the passenger</h3>

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

m is mass of the passengers
v1 is velocity of the first passenger
v2 is velocity of the second passenger

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

Learn more about momentum here: brainly.com/question/7538238

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5 0

2 years ago

A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t

OverLord2011 [107]

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

$(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s$

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

$s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s$

Now, the distance covered by the player in this time will be:

$s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)$

s₁ = 0.022 m

4 0

3 years ago

An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-

maw [93]

Answer:

Power requirement P for the banner is found to be 30.62 W
Power requirement P for the solid flat plate is found to be 653.225 W
Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The Cd for banner was given, whereas the Cd for a flat plate is generally found to be around 1.28 which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
Power requirement P for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
The power can be determined this equation: P = F . v, where F represents the drag-force that we will need to determine and v represents the velocity of the airplane
The equation to determine drag-force is: $F = 1/2 * d * C_d * A$

In the drag-force equation Cd represents the co-efficient of drag and A represents the frontal area of the banner/plate/balloon (the object being towed)

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

Part a) We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = 1/2 * 0.06* 1.225 * 20 = 0.735 N. Now to find the power P we will use P = F . v i.e. 0.735 * 41.66 = 30.62 W

Part b) For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = 1/2 * 1.28 * 1.225 * 20 = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = 653.225 W

Part c) The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

Part d) The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : pi* r^2 (r = d / 2). Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = 40.08 W

4 0

3 years ago

What is velocity and speed

hammer [34]

Velocity is height , speed is an object being propelled by force

5 0

3 years ago

Read 2 more answers

When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.

3 0

3 years ago