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natta225 [31]
3 years ago
13

True or False:

Physics
2 answers:
viva [34]3 years ago
8 0

Answer:

True. The object's location is based off of the reference point, so by moving the reference point, you move the object's location.

Explanation:

pantera1 [17]3 years ago
3 0

Answer:

True

Explanation:

the reference point is where the object starts so changing where it starts changes the objects location.

You might be interested in
A force of 20. Newtons to the left exerted on a cart for 10. Seconds. For what period of time must a 50.-newton force to the rig
FinnZ [79.3K]
Impulse = (force) x (time)

The first impulse was (20 N) x (10 sec) = 200 meters/sec

The second one is (50 N) x (time) and we want it equal to the first one, so

(50 N) x (time) = 200 meters/sec

Divide each side by 50N :    Time = 200/50 = <em>4 seconds</em>

By the way, the quantity we're playing with here is the cart's <em>momentum</em>.
6 0
3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
A system initially has an internal energy U of 504 J. It undergoes a process during which it releases 111 J of heat energy to th
Likurg_2 [28]

Answer:

615 J

Explanation:

internal energy (U) = 504 J

heat lost (q) = 111 J = - 111 J (negative sign is because heat is lost)

work done = 222 J

what is the final energy in the system

total energy = final energy - initial energy

final energy = total energy + initial energy

where

initial energy = 504 J

total energy = 222 - 111 = 111 J

final energy = 504 + 111 = 615 J

6 0
3 years ago
A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re
White raven [17]
A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law. 

I= \frac{V}{R}

Where:
I = current (A)
V = Voltage (V) (emf)
R = Resitance (Ω)

Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
V = IR

However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

\frac{1}{R_{T}} =  \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}


\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059
\frac{1}{R_{T}} =0.0613

Get the reciprocal of both sides and divide:

R_{T} =  \frac{1}{0.0613} =16.32

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
V = IR
V = (1.8)(16.32)
V = 29.376V

The emf of the battery is 29.376V


B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
I= \frac{V}{R}
I= \frac{29.376}{65}
I= 0.45A

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
I= \frac{V}{R}
I= \frac{29.376}{25}
I= 1.18A
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
I= \frac{V}{R}
I= \frac{29.376}{170}
I= 0.17A

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.
4 0
3 years ago
Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.2cm and a standard deviation o
Svet_ta [14]

Answer:

97.5%

Explanation:

By the empirical rule (68-95-99.7),

  1. 68% of data are within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>
  2. 95% of data are within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
  3. 99.7% of data are within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 2<em>σ</em>

<em>σ </em> and <em>μ</em> are the standard deviation and the mean respectively.

From the question,

<em>μ</em> = 7.2 cm

<em>σ</em> = 0.38 cm

7.96 = 7.2 + (<em>n</em> × 0.38)

<em>n</em> = 2

Hence, 7.96 represents <em>μ </em>+ 2<em>σ</em>.

P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)

P(X < <em>μ</em>) is the percentage less than the mean = 50%.

P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>) is half of P(<em>μ </em>- 2<em>σ</em> < X < <em>μ </em>+ 2<em>σ</em>) = 95% ÷ 2 = 47.5%.

Considering this, for apples that are no more than 7.96 cm,

P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%

<em />

5 0
3 years ago
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