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3 years ago

At a certain location, a gravitational force with

2 answers:

8 0

The unit of gravitational field strength is Nkg^-1 ie N/kg because the formula for Gravitational field strength is g=F/m where F is the gravitational force and m is the mass, and so by process of elimination, the magnitude of the gravitational field strength at this location is:
5.0 N/kg

laiz [17]3 years ago

4 0

The correct answer is: Option (2) 5.0 N/kg

Explanation:

To find the magnitude of the gravitational field strength, you need the formula of weight, which is:

weight = mass * gravitational-field-strength

Where weight is in Newtons, mass is in kilograms and gravitational-field-strength is (Newtons per kilogram).

In this particular scenario, the gravitational force is equal to the weight; please do not confuse weight with a mass. Mass is 70 kg.

Plug in the values:

350 = 70 * gravitational-field-strength

gravitational-field-strength = 350/70

gravitational-field-strength = 5.0 N/kg (Option 2)

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Red, green, and blue light rays each enter a drop of water from the same direction.Which light ray's path through the drop will

geniusboy [140]

Blue light will bend more than the others because it has a slightly greater refractive index. This is because blue light has a shorter wavelength and more energy, meaning it has to slow down more than the others when it hits the water.

5 0

3 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively