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2 years ago

A dog runs 3 miles and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Physics

1 answer:

Inga [223]2 years ago

6 0

Complete question:

A dog runs 3 miles east and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Answer:

The total distance covered by the dog is 7 miles

The displacement of the dog is 1 mile west

Explanation:

Given;

initial position of the dog = 3 miles east

final position of the dog = 4 miles west

time of motion, t = 6 hours

The total distance covered by the dog is given as;

Total distance = 3 miles + 4 miles = 7 miles

The displacement of the dog is given as;

displacement = final position of the dog - initial position of the dog

displacement = 4 miles west - 3 miles east = 1 mile west

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A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

Answer:

Maximum height reached = 35.15 meter.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

7 0

2 years ago

A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a

o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet $m=5 gm$

mass of wood block $M=1 kg$

Length of string $L=2 m$

Center of mass rises to an height of $0.38 cm$

initial velocity of bullet $u=450 m/s$

let $v_1$ and $v_2$ be the velocity of bullet and block after collision

Conserving momentum

$mu=mv_1+Mv_2$ -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

$\frac{Mv_2^2}{2}=Mgh_{cm}$

$v_2=\sqrt{2gh_{cm}}$

$v_2=\sqrt{2\times 9.8\times 0.38}$

$v_2=0.272 m/s$

substitute the value of $v_2$ in equation 1

$5\times 450=5\times v_1+1000\times 0.272$

$v_1=395.6 m/s$

4 0

2 years ago

(YOU WILL GET BRAINLIEST)Matter may be classified as a pure substance or a mixture. Where on the Venn diagram would you insert t

inessss [21]

I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms

3 0

2 years ago

Read 2 more answers

A student hears a police siren. What would change the frequency that the student hears? Check all that apply.

ser-zykov [4K]

A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source.

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.

4 0

2 years ago

Read 2 more answers

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th

zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(d) K = 11.8835 J

U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)

⇒ f = 1.3368 Hz

we use the formula

E = m*ω²*A² / 2

⇒ E = (1.15)(8.40)²(0.650)² / 2

⇒ E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)

and

U = (1/2)*m*ω²*x² (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒ K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒ U = 5.2581 J

4 0

2 years ago