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3 years ago

Which of the following is a vector quantity?

2 answers:

6 0

Since vectors are essentially quantities that have direction, D must be the answer as it contains a direction (down) while the other ones do not.

otez555 [7]3 years ago

3 0

A vector is a quantity with a size and a direction.

The only choice that has both of those is D .

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Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with

Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

½ m v₀² + m g h = ½ m v²

v² = v₀² + 2 g h

Let's calculate

v = √(1.23² + 2 9.8 1.69)

v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

W = ΔK

-fr x = 0- ½ m v²

Newton's second law

N- W = 0

The equation for the friction is

fr = μ N

fr = μ m g

We replace

μ m g x = ½ m v²

x = v² / 2μ g

Let's calculate

x = 5.89² / (2 0.255 9.8)

x = 6.94 m

6 0

3 years ago

Read 2 more answers

I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i

Lilit [14]

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

= ½ (10 m/s2) (4 s)2

= (5 m/s2) (16 s2)

= 80 ms

75% sure

Hope this helps!!!

5 0

3 years ago

A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

Ivahew [28]

<span>This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.</span>

5 0

3 years ago

Gauges all very according the vehicle , model make and year and other factors. By turning the vehicle to the ___ position you ca

AleksandrR [38]

Answer:

By turning the vehicle "ON" position you can check to see if the gauges light works.

When we switch ON or turn a key to ON the engine, we can find all the gauges working or not.

4 0

3 years ago