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3 years ago

Which of the following is an example in which you are traveling at constant speed but not at constant velocity?

Physics

1 answer:

grin007 [14]3 years ago

3 0

Answer:

c) Driving around in a circle at exactly 100 km/hr.

This examples represents constant speed but not constant velocity.

Explanation:

To answer this question it is importan to know the difference between the concept of speed and velocity.

speed: Is the ratio of change in the displacement per unit of time. JUST A MAGNITUDE

velocity: Is the ratio of change in the displacement per unit of time in a given direction. It is a composed value by magnitude and direction, this is know as a VECTOR.

So, in conclusion the speed is the magnitude or the scalar value, for the velocity vector.

In the example,

in a) Rolling freely down a hill in a cart, traveling in a straight line. There is a component of acceleration, from earth gravity, so the speed is changing.

in b) Driving backward at exactly 50 km/hr. The direction and speed are the same, so speed and velocity are constant.

In d) Jumping up and down, with a period of exactly 60 hops per minute. There are changes in the speed, due to the acceleration and decceleration between the changes in the direction. Actually you have to stop to change direction between the ups and downs in the jumps. Also there is the gravitational component, always changing the speed.

In c) Driving around in a circle at exactly 100 km/hr. In this case the speed remains constant, while the direction is changing all the time.

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2. Is this chemical equation balanced? 2C4H10 (g) + 13O2 (g) = 8CO2 (g) → 10H2O (l)

ehidna [41]

Yes that is a balaned equation

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3 years ago

A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis

GarryVolchara [31]

Resultant displacement is 29.2 km at $83.1^{\circ}$ north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

$A_x = 25.5 km\\A_y = 0 km$

- Second displacement is 41.0 km northwest, so its components are

$B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km$

So, the components of the resultant displacement are

$R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km$

And so, the magnitude is calculated using Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km$

And the direction is given by

$\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}$

Where the angle is measured from the west direction, since Rx is negative.

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

3 0

3 years ago

An object has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s.

taurus [48]

For the object that has a horizontal velocity component of 6 m/s and a vertical velocity component of 4 m/s we have:

1. The velocity of the object is 7.21 m/s.

2. The angle it makes with the horizontal is 33.7°.

1. The velocity of the object can be found as follows:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

Where:

$v_{x}$ : is the horizontal component of the velocity = 6 m/s

$v_{y}$ : is the vertical component of the velocity = 4 m/s

Hence, the velocity is:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(6 m/s)^{2} + (4 m/s)^{2}} = 7.21 m/s$

2. The angle it makes with the horizontal can be calculated with the following trigonometric function:

$tan(\theta) = \frac{v_{y}}{v_{x}}$

Where:

θ: is the angle it makes with the horizontal

Therefore, the angle is:

$\theta = tan^{-1}(\frac{v_{y}}{v_{x}}) = tan^{-1}(\frac{4}{6}) = 33.7$

You can learn more about the components of the velocity here: brainly.com/question/2285233?referrer=searchResults

I hope it helps you!

7 0

3 years ago

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibilty can be described by Hooke

Varvara68 [4.7K]

k = 1.7 x 10^5 kg/s^2 Player mass = 69 kg Hooke's law states F = kX where F = Force k = spring constant X = deflection So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration. F = kX F/X = k 115 kg* 9.8 m/s^2 / 0.65 cm = 115 kg* 9.8 m/s^2 / 0.0065 m = 1127 kg*m/s^2 / 0.0065 m = 173384.6154 kg/s^2 Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2 Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So X/0.39 cm = 115 kg/0.65 cm X = 44.85 kg/0.65 X = 69 kg The player masses 69 kg.

4 0

3 years ago

Read 2 more answers

Estimate the kinetic energy (in GJ) of a 93,000 metric ton aircraft carrier moving at a speed of at 32 knots. You will need to l

Wewaii [24]

Answer:

kinetic Energy = 12.58 GJ

Explanation:

1 metric ton is equal to 1000 kg

then,

93000 metric ton

Mass is m = 93000 x 1000 kg

speed is v = 32 knots

1 knot is 0.514 m/s

then ,

v = 32 x 0.514 = 16.448 m/s

To solve for the Kinetic Energy (KE), we have;

KE = 0.5 x m x v²

KE = 0.5*93000*1000*(16.448)²

= 12.58 x $10^{9}$ J

= 12.58 GJ

6 0

3 years ago