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NISA [10]
3 years ago
8

How does a arrow represent displacement

Physics
1 answer:
Lera25 [3.4K]3 years ago
4 0

Answer:

An arrow represents displacement. Because it's a vector quantity.

Vector quantity= Direction+Magnitude

When an arrow is put above a unit like

  • acceleration
  • displacement

it means that it's a vector quantity.

Hope this helps ;) ❤❤❤

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A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 30.0 min
schepotkina [342]

Answer:

Average speed of the trip = 52.9 km/h

Distance between initial pairs of cities (start to end) = 70.0 km

Explanation:

Since distance = speed × time

If she drives 30.0 min at 80.0 km/h

Distance covered = (30/60) × 80 = 40.0 km

Again she drives 45.0 min at 40 km/h

Distance covered = (45/60) × 40 = 30.0 km

Again she drives 12.0 min at 100 km/h

Distance covered = (12/60) × 100 = 20.0 km

Total distance covered =  40.0 + 30.0 + 20.0

                                       = 90.0 km

Total time spent = 30.0 + 45.0 + 12.0 +15.0

                           = 102 min

Average speed for the trip = Total distance covered/total time spent

                                             = 90/(102/60)

                                             = 52.9 km/h

Distance between initial cities will be between the start of one city to the end of another

Between the first pairs = 40.0 + 30.0 = 70.0 km

Between the second pairs = 30.0 + 20.0 = 50.0 km

6 0
3 years ago
A small piece of solid matter traveling through space is called a __________. When it collides with the atmosphere it is called
Oksanka [162]

Answer:

Meteoroid , meteor and meteorite

Explanation:

A meteoroid is a small rocky or metallic body travelling through space.  Most are fragments from comets or asteroids

The visible streak of light from space debris is the result of heat as it enters a planet's atmosphere, and the trail of glowing particles that it sheds in its wake is called a meteor, or colloquially a "shooting star" or "falling star". A series of many meteors appearing seconds or minutes apart, and appearing to originate from the same fixed point in the sky, is called a meteor shower. Incoming objects larger than several meters (asteroids or comets) can explode in the air. If a meteoroid, comet or asteroid or a piece thereof withstands ablation from its atmospheric entry and impacts with the ground, then it is called a meteorite.

8 0
3 years ago
A sphere of radius R = 0.295 m and uniform charge density -151 nC/m^3 lies at the center of a spherical, conducting shell of inn
cupoosta [38]

Answer:

a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C

Explanation:

a) r = 0.76R

As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:

Q = ρ*V(r) = ρ*\frac{4}{3} *\pi *r^{3}

where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³

Q = -151e-9 *\frac{4}{3} *\pi *0.224m^{3} = -7.11e-9 C

Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:

E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-7.11e-9C}{(0.76*0.295m)^{2}} =-1.27e3 N/C

⇒ E = -1.27*10³ N/C

b) r= 3.90 R

As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0

c) r = 2.8 R

As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.

First we need to find the total charge of the sphere, as follows:

Q = ρ*V =

Q = -151e-9 *\frac{4}{3} *\pi *0.295m^{3} = -16.2e-9 C

In the same way that for a) we can write the following expression:

E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-16.2e-9C}{(2.8*0.295m)^{2}} =-0.21e3 N/C

⇒ E = -0.21*10³ N/C

d) r= 7.30 R

In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.

As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.

This leaves an excess charge on the outer surface of the shell as follows:

Qsh = 66.7 nC - 16.2 nC = 50.5 nC

Now, we can repeat the same process than for a) and c) as follows:

E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{50.5e-9C}{(7.3*0.295m)^{2}} =0.1e3 N/C

⇒ E = 0.1*10⁻³ N/C

6 0
2 years ago
New york to Los Angeles is roughly 2,800 miles. If you drive 70mph, how long will it take to travel that distance?
Gelneren [198K]

2800 Miles / 70 Miles an hour = 40 hours.

7 0
3 years ago
HELP ME PLSSS What are the possible genotypes
riadik2000 [5.3K]

Answer:

the answer is b i think

7 0
2 years ago
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