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3 years ago

Can the magnitude of a vector have a negative value? Explain.

1 answer:

8 0

The magnitude of a vector is the square root of its components squared. When you square a negative number, you will always get a positive number. Its like speed of your car. It will always be positive (or 0) whether you are moving forward or you are in reverse gear or stationary.

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A horizontal 791-N merry-go-round is a solid disk of radius 1.56 m and is started from rest by a constant horizontal force of 49

ICE Princess25 [194]

Answer:

$K.E=273.5J$

Explanation:

Given data

$F_{g}=791N\\F_{h}=49N\\r=1.56m\\t=3.03s$

To find

Kinetic Energy

Solution

The moment of inertia is given as:

$I=(\frac{1}{2} )MR^2\\I=\frac{1}{2}(F_{g}/g)R^2\\ I=\frac{1}{2}(\frac{791N}{9.8m/s^2} )(1.56m)^2\\ I=98.21kg.m^2$

The angular acceleration is given as:

$\alpha =\frac{T}{I}\\\alpha =\frac{F_{y}R}{I}\\ \alpha =\frac{(49N)(1.56m)}{98.2kg.m^2}\\\alpha =0.778rad/s^2$

Now the angular velocity is given by:

$w=\alpha t\\w=(0.778rad/s^2)(3.03s)\\w=2.36rad/s$

So the kinetic energy given as:

$K.E=(\frac{1}{2} )Iw^2\\K.E=\frac{1}{2}(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J$

8 0

3 years ago

A 1,000 kg ball traveling at 5 m/s would have

jonny [76]

Answer:

15 because 5×5×5 is the same thing as 5×3 which equals to 15

6 0

3 years ago

Read 2 more answers

A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley c

sergiy2304 [10]

Answer:

Part a)

$a_{hanger} = g - \frac{T}{m}$

Part b)

$a_{cart} = \frac{T}{M}$

Part c)

$a = \frac{mg}{M + m}$

Part d)

$a = 1.35 m/s^2$

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

$mg - T = ma$

so we have

$a_{hanger} = g - \frac{T}{m}$

Part b)

now for car that is rolling on the floor the net force is given as

$F = Ma$

$T = Ma$

$a_{cart} = \frac{T}{M}$

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

$T = Ma$

$mg - Ma = ma$

$a = \frac{mg}{M + m}$

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

$a = \frac{0.50(9.81)}{2.40 + 0.50}$

$a = 1.35 m/s^2$

6 0

3 years ago

A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s a

arsen [322]

Answer:

The apparent weight is 5 times greater than the original weight at the bottom.

Explanation:

Given:

Mass of the pilot, m = 84 kg

Velocity of the jet, v = 345 m/s

Radius of the loop, R = 3.033 km = 3.033 * 10^3 m

We have to find the apparent weight that the pilot feels.

Let the apparent weight be "N" .

Apparent weight :

It is based on where is the position of the pilot in the loop-the-loop.
The apparent weight is the highest at the bottom of the loop-the-loop.
Because the weight acts down and the normal force acts towards the center of the circle.

From the FBD shown we can say that :

apparent weight (N)

⇒ $N=mg+\frac{mv^2}{R}$

⇒ $N=mg(1+\frac{v^2}{Rg} )$

⇒ $N=mg(1+\frac{(345)^2}{3.033\times 10^3 \times 9.8} )$

⇒ $N=5mg$

Therefore,

The force exerted by the seat on the pilot at the bottom of the loop is greater than the pilots weight by a factor of 5.

7 0

3 years ago

Local action in a simple cell may be overcomed by

sergeinik [125]

Zinc plate used in the simple cell is not pure so impurities present in it provides a conducting path for the flow of charge. This makes zinc wasted. It is called the local action. It can be removed by using pure zinc or amalgamated zinc rod.

6 0

3 years ago