All categories

3 years ago

A car travels 35 km west and 90 km north in two hours what is its average velocity?

1 answer:

8 0

Average velocity = Total Displacement / Total time taken

Total displacement is the distance from starting to end point.

35km west then 90km north. When you will draw this and connect starting to end point. You will get a right angled triangle.

So displacement = $\sqrt{35^2+90^2}$

= 96.56 km

So, Total displacement = 96.56 km

So, Average Velocity = 96.56 / 2 = 48.28 km/hr

You might be interested in

What are the conditions to increase the stability of a body?

cricket20 [7]

Training everyday
..........

5 0

3 years ago

Sound waves travel fastest through a A) gas. B) liquid. C) solid. D) vacuum.

earnstyle [38]

Sound waves travel faster through solids than they do through gases or liquids. (C) They don't travel through vacuum at all.

Example:

Speed of sound in normal air . . . around 340 m/s

Speed of sound in water . . . around 1,480 m/s

Speed of sound in iron . . . around 5,120 m/s

3 0

3 years ago

Read 2 more answers

Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni

Fudgin [204]

Answer:

$|\vec A + \vec B| = 50 units$

Explanation:

As we know that magnitude of two vectors is given as

$|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}$

here we know that

A = magnitude of vector A

B = magnitude of vector B

$\theta$ = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

$|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}$

$|\vec A + \vec B| = \sqrt{30^2 + 40^2}$

$|\vec A + \vec B| = 50 units$

3 0

3 years ago

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95

madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as y=l'-l=41-30=11m

the mass is m=95 kg

the force is F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of k is given by equating the energy at the equilibrium point which is given as

$U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}$

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

$k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m$

Now the force is

$F=kx\\$ or

$x=\dfrac{F}{k}\\$

So here F=380 N, k=630.92 N/m

$x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m$

So the distance is 0.602 m

6 0

3 years ago

The energy of a given wave in the electromagnetic spectrum is 2.64 × 10-21 joules, and the value of Planck’s constant is 6.6 × 1

bixtya [17]

We are given
E = 2.64 × 10-21 J
h = 6.6 × 10-34 J s

The options given below are frequencies, therefore, the question must be asking about the frequency fo the given wave

The equation is
E = h f
Simply substitute and solve for f which is the frequency
f = 2.64 × 10-21 J / 6.6 × 10-34 J s
f = 4.00 × 1012 hertz

6 0

3 years ago