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noname [10]
2 years ago
13

Use the fact that 1inch=2.54cm and convert 52 inches into the equivalent length in centimeters.

Physics
1 answer:
Lelu [443]2 years ago
4 0

Answer:

To convert inches to centimeters, use an easy formula and multiply the length by the conversion ratio.

Since one inch is equal to 2.54 centimeters, this is the inches to cm formula to conver

Explanation:

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The force of gas particles against the walls of a container is called ________
Liono4ka [1.6K]

Answer:

the answer is pressure

8 0
3 years ago
How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?
4vir4ik [10]

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

5 0
3 years ago
The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo
igomit [66]

Suppose car A is moving with a velocity Va, and car b with a velocity Vb,

According the principle of conservation of momentum:

Va x Ma + Vb x Mb = (Ma + Mb) V

V = (Va x Ma + Vb x Mb)/(Ma +Mb)

V = speed of cars after coupling

V = (Va x 20 mg + Vb x  15 mg)/(20 mg + 15 mg)

Put in the values of Va and Vb, and get the V

7 0
3 years ago
Read 2 more answers
A space shuttle takes off from FL and circles Earth several times, finally landing in CA. While the shuttle is in flight, a phot
mixer [17]

Answer:

Both the astronauts and photographer have the same displacement

Explanation:

Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.

Hence, both the astronauts and photographer will have the same displacement.

3 0
4 years ago
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
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