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2 years ago

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Detlev walks 1000 meters north to the store and walks back 900 meters south to his friends house. The friends house is 100 meter

s to the north of Detlev’s house. What was Detlev’s Travel Distance?

1 answer:

lilavasa [31]2 years ago

3 0

1000 is the correct answer

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Jonah observes the Sun through a special filtered telescope during a total solar eclipse. He sees a red ring and a faint white r

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The answer would be C. chromosphere and corona.

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3 years ago

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the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i

Elanso [62]

Gravitational potential energy can be calculated using the formula:

$PE_{grav} =mgh$

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it.
$PE_{grav} =mgh$
$PE_{grav} =(10kg)(3.7m/s^{2})(1m)$
$PE_{grav} =37J$

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3 years ago

Pleaseeee help I really need to pass to graduate

Rus_ich [418]

Answer:

kWh I think

Explanation:

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3 years ago

Two concentric, coplanar, circular loops of wire of different diameter carry currents in the same direction. Describe the nature

Ierofanga [76]

Answer:

Both will be attractive in nature.

Explanation:

In the given case, the direction of the magnetic field is same in both loops as the direction of the current is same in both loops. When two parallel straight wire carrying current in the same direction are brought close to each other, the force between them is attractive in nature. In the same way, when two coplanar, circular and concentric loops of wire are carrying current in the same direction, the force between them is attractive in nature. It can be checked by using right hand thumb rule.

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2 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

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2 years ago