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sveticcg [70]
3 years ago
14

What is the word that goes with the definition “ a disturbance in space that causes a force”

Physics
1 answer:
irga5000 [103]3 years ago
8 0

Answer:

Gravity

Explanation:

Gravity is the force that causes a "disturbance" in space between planets.

You might be interested in
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
GREYUIT [131]

Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

6 0
3 years ago
This is for physical science. Can someone help me?
Radda [10]
A. 320 g
B. 160 g
C. 80 g
D. 40 g
6 0
3 years ago
Read 2 more answers
A ball droped from a building. How fast is it traveling after falling 3.55s
algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

d=ut+\dfrac{1}{2}at^2

Here, u = 0 and a =g

d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m

So, it will cover 61.75 m after falling 3.55 seconds.

4 0
3 years ago
Can anyone please help me with the steps
KIM [24]

Answer:

A) v_average = - 10 km / h,  B)    v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

    v₁ = 40 km / h

    v₂ = - 60 km / h

the average speed is

         v_average = \frac{v_1 + v_2}{2}

         v_average = ( 40 - 60)/2

         v_average = - 10 km / h

B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

i) the speed for 2.5 s above

         v = v₀ + a t

as the time is earlier

          t = - 2.5 s

we substitute

          v = 9.6 - 3.2 2.5

          v = 1.6 m / s

ii) the velocity for a subsequent time of 2.5 s

          t = 2.5 s

           

we substitute

          v = 9.6 + 3.2 2.5

          v = 17.6 m / s

3 0
3 years ago
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