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slava [35]
4 years ago
13

If 1.0 joule of work is required to move a charge of 1.0 coulomb between two points is an electric field the potential differenc

e between these two point is?
Physics
1 answer:
deff fn [24]4 years ago
6 0

Answer:1V

Explanation: simply put it

V=W/q

=1/1

=1V

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What do parentheses mean in a Calc formula? A. Ignore the numbers in parentheses until the end. B. Calculate the numbers in pare
kirill115 [55]
The answer is B.........
3 0
3 years ago
Read 2 more answers
Car B has the same mass as Car A but is going twice as fast. If the same constant force brings both cars to a stop, how far will
evablogger [386]

Answer:four times

Explanation:

Given

mass of both cars A and B are same suppose m

but velocity of car B is same as of car A

Suppose velocity of car A is u

Velocity of car B is 2 u

A constant force is applied on both the cars such that they come to rest by travelling certain distance

using  to find the distance traveled

where, v=final velocity

u=initial velocity

a=acceleration(offered by force)

s=displacement

final velocity is zero

For car A

0-(u)^2=2\times a\times s

s_a=\frac{u^2}{2a}------1

For car B

0-(2u)^2=2\times a\times s_b

s_b=\frac{4u^2}{2a}----2

divide 1 and 2 we get

\frac{s_a}{s_b}=\frac{1}{4}

thus s_b=4\cdot s_a

distance traveled by car B is four time of car A

7 0
3 years ago
Acceleration = change of velocity divided by time interval = Δv/Δt.
MariettaO [177]

Answer:

a=2.378 m/s^2

Explanation:

a=Δv/Δt------eq(1)

Δv=Vf-Vi=120 km/h-0 km/h=120 km/h

or Δv=33.3 m/sec

or time=t=14s

putting values in eq(1)

a=33.3/14

a=2.378 m/s^2

6 0
3 years ago
Please don't post anything unless it's the answer
Alja [10]
52800000000000000000000000000000000000000000
7 0
3 years ago
An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd
Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ)  β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

    = 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

    = 1.07 × 10³(1 - [0.024]²)

    =  1.07 × 10³(1 - 0.000576)

    = 1.07 × 10³(0.999424)

    = 1069.38 gallons

7 0
4 years ago
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