2 years ago

At which position would a person on earth see the entire lighted half of the moon?

1 answer:

3 0

The best position for the person would be outside, under a clear sky, standing up. He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.

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A bus travelling 30.0 km/h east has a constant increase in speed of 1.5 m/s. What is its velocity 6.8s later

alekssr [168]

Answer:

8.72 seconds before

Explanation:

and 27.1 secs after

7 0

2 years ago

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is $4.8 \cdot 10^{-6}\,mg$

Explanation:

The ration between Uranium mass and total sample mass is: $\frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}$

For a sample of mass 1.2 mg, the amount of uranium is:

$1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg$

7 0

2 years ago

A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ign

Gemiola [76]

B- the acceleration is greater for the more massive rock

5 0

1 year ago

Read 2 more answers

What are some solutions to global temperature changes?<br>help asap

Nimfa-mama [501]

Answer:

Power your home with renewable energy.

Weatherize, weatherize, weatherize.

Invest in energy-efficient appliances.

Reduce water waste.

Actually eat the food you buy—and make less of it meat.

Explanation:

6 0

2 years ago

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4

irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

$mvL = (I + mL^2)\omega$

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

$2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega$

$1.75 = [4.5 + 0.245]\omega$

$1.75 = 4.745\omega$

$\omega = 0.37 rad/s$

so the final angular speed will be 0.37 rad/s

4 0

2 years ago