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3 years ago

During a solar eclipse, which of the following is true?

Physics

2 answers:

Helga [31]3 years ago

8 0

Answer:

The Moon blocks the Sun's light from hitting the surface of the Earth.

kaheart [24]3 years ago

6 0

Answer:

The Moon blocks the suns light from hitting the surface of the earth

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The speed of water flowing through the 4in diameter section of the piping system is 3ft/s. What is the volume flow rate of water

olga_2 [115]

1 ft =12 in
4 in = 0.333 ft
volume = (п/4)*(0.333)² = 0.087 ft²
vol. flow = spead *volume
=3 ft/s * 0.087 ft²

vol flow = 0.261 ft³/s

3 0

3 years ago

Discuss impact of Newton's second law on roads

ruslelena [56]

Answer:

The second law: When a force is applied to a car, the change in motion is proportional to the force divided by the mass of the car. This law is expressed by the famous equation F = ma, where F is a force, m is the mass of the car, and a is the acceleration, or change in motion, of the car

4 0

2 years ago

What is a newton defined as<br>

slava [35]

Answer:

Newton, absolute unit of force in the International System of Units (SI units), abbreviated N. It is defined as that force necessary to provide a mass of one kilogram with an acceleration of one metre per second per second.

4 0

3 years ago

The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

4 0

3 years ago

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

7 0

3 years ago