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3 years ago

PLEASE ANSWER ALL! CORRECT ANSWERS ONLY!!!!

1 answer:

3 0

Distance is a scalar quantity. It refers only to how far an object has traveled. For example, 4 feet is a distance; it gives no information about direction. To say an object traveled 4 feet is somewhat ambiguous. To say the object traveled 4 feet west, for example, would be a displacement, and would then be a vector quantity. It gives a more complete picture of what happened.

Mass is a scalar quantity. Simply put, it refers to how much matter an object is made up of. It has magnitude but gives no indication of direction in any sense. The vector counterpart to mass is weight.

Weight is a vector quantity. Weight is a force, and forces are vectors, i.e. having both magnitude and direction. Perceived weight of an object at rest on earth is given by

→

, the product of the mass of the object and the free-fall Time is a scalar quantity (as far as we are concerned at this level). It gives information about magnitude, i.e. how much time, but no information about direction.

Volume is a scalar quantity. It refers to the amount of space that an object occupies and therefore has magnitude, but gives no information about direction.

Density is a scalar quantity, having only magnitude and giving no information about direction. We can also reason that, because density is equal to mass divided by volume and both mass and volume are scalar quantities, density must also be a scalar quantity.

Speed is a scalar quantity, having only magnitude and giving no information about direction. For example,

is a speed, it tells us how fast an object is traveling, but nothing abut which direction the object is traveling in. The vector counterpart to speed is velocity.

Velocity is a vector quantity. Velocities have both magnitude and direction. For example,

NE is a velocity.

Acceleration is a vector quantity. Acceleration has both magnitude and direction. For example,

9.8

downward or

−

9.8

vertically is an acceleration.

Force is a vector quantity. Force has both magnitude and direction. Weight is an example of force given above. Another is the force of friction, which has some magnitude and acts in the direction opposite that of motion.

Temperature is a scalar quantity. A measurement of temperature has magnitude, but gives no information about direction.Note that certain quantities which are alone scalars can be represented as vectors when we discuss intervals or how the quantities change (e.g. measure of increase or decrease). this is all you need to know my friend

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Two people are pushing a car to the right. The mass of the car is 1000.0 kg. One person applies a force of 275 N to the car, whi

givi [52]

Explanation:

Below is an attachment containing the solution.

7 0

2 years ago

A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i

scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

then q=-18.28 cm

The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.

6 0

3 years ago

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes thro

LiRa [457]

Answer:

$x = t$

$y = \frac{1}{3}t$

$z =t$

Explanation:

Given

$r(t) = f(t)i + g(t)j + h(t)k$ at $t = 0$

Point: $(f(t0), g(t0), h(t0))$

$r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk$ , $t0 = 1$ -- Missing Information

Required

Determine the parametric equations

$r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk$

Differentiate with respect to t

$r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k$

Let t = 1 (i.e $t0 = 1$ )

$r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k$

$r'(1) = i +\frac{3}{3^2}j + (0 + 1)k$

$r'(1) = i +\frac{3}{9}j + (1)k$

$r'(1) = i +\frac{1}{3}j + (1)k$

$r'(1) = i +\frac{1}{3}j + k$

To solve for x, y and z, we make use of:

$r(t) = f(t)i + g(t)j + h(t)k$

This implies that:

$r'(1)t = xi + yj + zk$

So, we have:

$xi + yj + zk = (i +\frac{1}{3}j + k)t$

$xi + yj + zk = it +\frac{1}{3}jt + kt$

By comparison:

$xi = it$

Divide by i

$x = t$

$yj = \frac{1}{3}jt$

Divide by j

$y = \frac{1}{3}t$

$zk = kt$

Divide by k

$z = t$

Hence, the parametric equations are:

$x = t$

$y = \frac{1}{3}t$

$z =t$

3 0

3 years ago

How would you determine the volume of an irregularly shaped object such as a paper clip?

zheka24 [161]

Explanation:

Take a measuring cylinder and fill it with a certain amount of water. Measure this amount of water.

Place the paper clip in the filled measuring cylinder. You will notice that the water level has gone up. When we place the paper clip in the cylinder the volume of the paper clip gets added to the volume that was present in the cylinder.

The volume of the paper clip will be the final volume of water with the paper clip - The initial volume of water without the paper clip.

Any irregularly shaped object's volume can be determined by this method.

3 0

3 years ago

A block slides down a frictionless incline with constant acceleration. After sliding 6.80 m down, it has a speed of 3.80 m/s. Wh

melisa1 [442]

Answer:

speed when the block had slid 3.40 m is 2.68 m/s

Explanation:

given data

distance = 6.80 m

speed = 3.80 m/s

to find out

speed when the block had slid 3.40 m

solution

we will apply here equation of motion that is

v²-u² = 2×a×s ..............1

here s is distance, a is acceleration and v is speed and u is initial speed that is 0

so put here all value in equation 1 to get a

v²-u² = 2×a×s

3.80²-0 = 2×a×6.80

a = 1.06 m/s²

speed when distance 3.40 m

from equation 1 put value

v²-u² = 2×a×s

v²-0 = 2×1.06×3.40

v² = 7.208

v = 2.68

so speed when the block had slid 3.40 m is 2.68 m/s

7 0

3 years ago