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Nitella [24]
3 years ago
7

Help physics! 1. Was the object at rest during the time interval labeled "B"?

Physics
2 answers:
Rina8888 [55]3 years ago
7 0

Answer:

Q1: No

Q2: No

Explanation:

the slope of velocity time graph gives the value of acceleration.

In section B

the graph is parallel to X axis, it means it is moving with constant velocity. Here the slope of graph in section B is zero.

Q1: Object is moving with constant velocity

Q2: The acceleration is zero

myrzilka [38]3 years ago
5 0
1) Yes 2) No, hope this helps...
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Jake drops a book out of a window from a height of 10 meters. At what velocity does the book hit the ground?
user100 [1]
G = 9.81 m/sec^2)     g = 9.81\frac{meters}{ second^{2} }

<span>Solving for velocity : </span>

velocity^{2}<span> = 2gh </span>
<span>v = </span>2gh^{ \frac{1}{2} }
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
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3 years ago
Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(
Marianna [84]

So that we do not contaminate it with microorganisms or garbage or other human stuff.

5 0
2 years ago
a vehicle travels at a constant speed of 65 mph for 4 hours how far has this vehicle travelled in this time
Nostrana [21]
260 miles.................
3 0
3 years ago
How much energy will a stock tank heater rated at 1790 Watts use in a 24 hour period? 1. 1790 × 24 × 3600 Joules 2. 1790 × 24 ×
Rashid [163]
<h2>Option 1 is the correct answer.</h2>

Explanation:

Power of heater, P = 1790 W

Time used, t = 24 hours = 24 x 60 x 60 = 24 x 3600 s

We have the equation

               \texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}

We need to find energy,

Substituting

                 \texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}\\\\1790=\frac{\texttt{Energy}}{24\times 3600}

                  Energy = 1790 x 24 x 3600 J

Option 1 is the correct answer.

3 0
2 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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