draw the following vector quantity Using the coordinate system.
a. 190 newton east
b. 120km/hr, 250 north of east
c. 60 meters southwest​

Answers

Answer 1

The given vectors quantities can be described by their properties of both

magnitude and direction.

a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.b. The velocity vector extending from  (0, 0) to (108.76, 50.714) on the coordinate plane is attached.c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

[tex]\vec{F}[/tex] = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; [tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j, which gives;

[tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

[tex]\vec{v}[/tex] ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is [tex]\vec{d}[/tex] = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

[tex]\vec{d}[/tex] = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

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Draw The Following Vector Quantity Using The Coordinate System. A. 190 Newton East B. 120km/hr, 250 North
Draw The Following Vector Quantity Using The Coordinate System. A. 190 Newton East B. 120km/hr, 250 North
Draw The Following Vector Quantity Using The Coordinate System. A. 190 Newton East B. 120km/hr, 250 North

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A wagon with a mass of 284 kg is accelerated across a level surface at 9 m/s2. What net force acts on the wagon?

Answers

Answer:

2556 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 284 × 9 = 2556

We have the final answer as

2556 N

Hope this helps you


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Answers

Answer:

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Answer:

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Explanation:

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Answers

Trigonometry allows to find the result for the question about the height of the mountain is:

The height mountain  is:  y = 9674.4 ft

Trigonometry allows finding relationships between the angles of a right triangle.

         [tex]tan \theta = \frac{y}{x}[/tex]  

Where θ is the angle, y the opposite leg (height) and x the adjacent leg (horizontal distance).

In the attachment we can see a diagram of the system. They indicate that for x  distance the angle is 14.67º  

         tan 14.67 = [tex]\frac{y}{x}[/tex]  

At the other point the angle is 10.38º.

        tan 10.38 = [tex]\frac{y}{x+15860}[/tex]

We look for the horizontal distance (x) with these equations.

        x tan 14.67 = (x + 15860) tan 10.38

        x tan 14.67 / tan 10.38 = x + 15860

        x 1,429 = x + 15860

        x 0.429 = 15860

        x = [tex]\frac{15680}{0.429}[/tex]

        x = 36955.3 ft

We calculate for the height.

        y = x tan 14.67

        y = 36955.3 tan 14.67

        y = 9674.4 ft

In conclusion using trigonometry we can find the result for the height of the mountain is:

The height is y = 9674.4 ft

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Energy conservation relies on people cutting back on activities that consume energy—by turning off lights or driving less or using appliances less often.

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Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a greater displacement?

Answers

Answer:

The distance of one lap on this track from the start line to the finish line is 400 meters. Two laps around the track is 800 meters, half a lap is 200 meters, and so on.

    Remember, displacement is the direction from the starting point and the length of a straight line from the starting point to the ending point. If a runner was going to run 400 meters (one lap around the track), they would start and stop at the same place on the track. Therefore, their displacement would be 0.

 

Explanation:

hope this helped

Answer:

[tex]\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}[/tex]

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has no displacement. If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has 50 meters of displacement. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, Gabi had the greater displacement.

A pail of water of mass 2000 gm is rotated in a vertical circle of radius 1.00 m. What is the pail’s minimum speed at the top of the circle if no water is to spill out and the tension exerted by the string is 25 N?

Answers

The pail’s minimum speed at the top of the vertical circle is 4.72 m/s.

The given parameters:

Mass of the water, m = 2000 g = 2 kgRadius of the circle, r = 1 mTension on the string, T = 25 N

The net force on the pail of water at the top of the vertical circle is calculated as follows;

[tex]T = ma - mg\\\\T = \frac{mv^2}{r} - mg\\\\T + mg = \frac{mv^2}{r} \\\\mv^2 = r(T + mg)\\\\v^2 = \frac{r(T + mg)}{m} \\\\v= \sqrt{\frac{r(T + mg)}{m}} \\\\v = \sqrt{\frac{1 (25\ + \ 2 \times 9.8)}{2}} \\\\v = 4.72 \ m/s[/tex]

Thus, the pail’s minimum speed at the top of the vertical circle is 4.72 m/s.

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The pail’s minimum speed at the top of the circle if no water is to spill out is; v = 4.722 m/s

We are given;

Mass of the pail of water; m = 2000 g = 2kg

Radius of the circle; r = 1 m

Tension exerted by string; T = 25 N

To calculate the net force on the pail of water at the top of the vertical circle, let us take equilibrium of forces to give;

W - T = ma

Where;

W is weight = mg

T is tension on the string

Now, in circular motion, we know that;

Acceleration is; a = v²/r

Thus;

mg - T = m(v²/r)

Making v the subject of the formula gives us;

v = √[(r(T + mg)/m)

Plugging in the relevant values gives;

v = √[(1(25 + (2 * 9.8))/2)

v = 4.722 m/s

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Suppose you drop a 10-pound weight and a 5-pound weight on the Moon, both from the same height at the same time. What will happen? The 10-pound weight will hit the ground before the 5-pound weight. The 5-pound weight will hit the ground before the 10-pound weight. Both weights will float freely because everything is weightless on the Moon. Both will hit the ground at the same time.

Answers

Answer:

Both will hit the ground at the same time.

batman recharges his bat taser with 2500 joules of energy in 15 minutes. how much power does it draw from the mains

Answers

The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
P(W) = E(J) / t(s)

2.78 Watts

A 30.4-newton force is used to slide a 40.0-newton crate a distance of 6.00 meters at constant speed along an incline to a vertical height of 3.00 meters. Calculate the work done by a friction force on the crate as it slides 6.00 m along the incline.

Answers

Hi there!

Since the crate is being slid at a constant speed, the forces sum to 0 N. In this instance, the following forces occur in the axis of interest:

Wsinθ = downward acceleration along incline due to gravity (N)

Fκ = kinetic friction force along incline (N)

A = applied force (N)

The acceleration due to gravity and friction force act in the same direction, so:

Wsinθ + Fκ = A

Solve for sinθ using right triangle trigonometry:

sinθ = O/H = 3/6 = 0.5

Rearrange the equation for the force of kinetic friction and solve:

Fκ = A - 0.5W

Fκ = 30.4 - 20 = 10.4 N

Now, recall that:

Work = Force × displacement (W = F × d)

Since the box's displacement is in the same axis as the force but OPPOSITE direction, we must use:

W = Fdcosθ

Angle between displacement and friction force is 180°.

cos(180) = -1

Work done by friction = -Fd = -10.4(6) = -62.4 J

Anyone please.??????​

Answers

Answer:

468.42572 Wavelength In Metres

Explanation:

Answer:

[tex]\huge\boxed{\sf Wavelength= 470\ m}[/tex]

Explanation:

Given Data:

Speed of light = c = 3 × 10⁸ m/s (Constant)

Frequency = f = 640 kHz = 6.4 × 10² × 10³ Hz = 6.4 × 10⁵ Hz

Required:

Wavelength = λ = ?

Formula:

λ = c / f

Solution:

λ = 3 × 10⁸ / 6.4 × 10⁵

λ = 0.47 × 10³

λ = 4.7 × 10² m

λ = 470 m

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

~AH1807
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