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3 years ago

11

A vector has what and direction

2 answers:

galben [10]3 years ago

8 0

A vector Quantity has MAGNITUDE and direction, so the ‘what’ is Magnitude.

lys-0071 [83]3 years ago

4 0

Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. Although a vector has magnitude and direction, it does not have position.

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A SMALL cup of tea has water inside that is very hot, about 80 degrees Celsius. A LARGE bathtub has water inside it that is warm

Artist 52 [7]

Answer:

the water in the tea cup

Explanation:

Kinetic energy is the energy an object or particles possesses as a result of motion.

Liquid in higher temperature possesses more kinetic energy than liquid with lower temperature

7 0

3 years ago

A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?

taurus [48]

Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

$V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]$

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

8 0

2 years ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

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8 0

1 year ago

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

Citrus2011 [14]

Answer:

The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

$\tau=I\times\alpha$

Here, I = mk²

$\tau=mk^2\times\alpha$

Put the value into the formula

$\tau=3\times(0.2)^2\times0.5$

$\tau=0.06\ N-m$

$\tau=6.0\times10^{-2}\ N-m$

Hence, The magnitude of the applied torque is $6.0\times10^{-2}\ N-m$

7 0

3 years ago

A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of

bulgar [2K]

m = mass of the car moving in horizontal circle = 1750 kg

v = Constant speed of the car moving in the horizontal circle = 15 m/s

r = radius of the horizontal circular track traced by the car = 45.0 m

F = magnitude of the centripetal force acting on the car

To move in a circle . centripetal force is required which is given as

F = m v²/r

inserting the above values in the formula

F = (1750) (15)²/(45)

F = (1750) (225)/(45)

F = 1750 x 5

F = 8750 N

6 0

3 years ago