All categories

2 years ago

A dog runs 30 feet to the north then 5 feet to the south what is the displacement of the dog

Physics

2 answers:

Nuetrik [128]2 years ago

5 0

To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.

RUDIKE [14]2 years ago

4 0

The initial displacement=d1= 30 feet North

d2= -5 feet south

we take d2 as negative because we treat south direction as negative

so the net displacement= d1+d2

net displacement= -5+30

net displacement=25

so the net displacement= 25 feet North.

You might be interested in

Helium (density 0.18 kg/m^3 at 0°C and 1 atm pressure) remains a gas until the extraordinarily low temperature of 4.2 K.What is

balu736 [363]

Answer:

$166.6396m/sec$

Explanation:

Molar mass of helium = $4\times 10^{-3}kg/mole$

$\gamma$ for helium is 1.67

The velocity of helium in sound at any temperature is given by $v=\sqrt{\frac{\gamma RT}{M}}=\sqrt{\frac{1.67\times 8.314\times 8}{4\times 10^{-3}}}=166.6396m/sec$

R is a constant 8.314 $atm\ mol^{-1}K^{-1}$

7 0

2 years ago

After soccer practice, Coach Miller goes to the roof of the school to retrieve the errant soccer balls. The height of the school

blsea [12.9K]

With gravitational acceleration at 9.8, initial height at 3.5m and distance at 22m the initial horizontal velocity is 26.03 ms and the flight time is .845 seconds

3 0

2 years ago

A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards

Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0

2 years ago

A student lifts a 50 pound (lb) ball 4 feet (ft) in 5 seconds (s). How many joules of work has the student completed?

Elis [28]

   Work = (weight) x (distance)

Work = (50 lb) x (1 kg / 2.20462 lb) x (9.81 newton/kg)

   x (4 feet) x (1 meter / 3.28084 feet)

   = (50 x 9.81 x 4) / (2.20462 x 3.28084) newton-meter

   =    271.3 joules .

We don't need to know how long the lift took, unless we
want to know how much power he was able to deliver.

   Power = (work) / (time)

   = (271.3 joule) / (5 sec) = 54.3 watts .
________________________________________

The easy way:

   Work = (weight) x (distance)

   = (50 pounds) x (4 feet) = 200 foot-pounds

Look up (online) how many joules there are in 1 foot-pound.

There are 1.356 joules in 1 foot-pound.

So 200 foot-pounds = (200 x 1.356) = 271.2 joules.

That's the easy way.

5 0

2 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

2 years ago