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3 years ago

A book has a mass of 50 grams and an acceleration of 5 m/s/s. What is the force on the book?

Physics

2 answers:

faust18 [17]3 years ago

4 0

Answer:

250 N ( Newtons)

Explanation:

mass X acceleration = F net or in this case Force

kg X m/s/s = N

IRISSAK [1]3 years ago

4 0

From second law of motion, we have,

F = ma

So, F = 50kg × 5m/s^2 = 250 kg m s^-2 = 250 N

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A person approaches a flat mirror at a speed of 0.25 m / s. How fast is he approaching his image?

Bezzdna [24]

Flat mirror is given here
We know that in a flat mirror the distance of object from the mirror is equal to distance of image from the mirror i.e v=u
Only the side of image changes ime left is seemed right.

So the speed remains same 0.25m/s.

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3 years ago

Neurons that relay information within the cns and are the location of information processing are called __________.

emmainna [20.7K]

Answer:

Interneurons

Explanation:

An interneuron or integrative neuron is a central nervous system neuron, usually small and short axon, that interconnects with other neurons; but never with sensory receptors or muscle fibers, allowing more complex functions.

The interneuron, also called the association neuron, has the function of analyzing sensory information and storing part of it. It also acts on reflex acts, transforming a stimulus in response at the level of the spinal cord. They are located between sensory and motor neurons and are located in the upper nerve centers. Interneurons are multipolar neurons, which connect afferent neurons with efferent neurons in the neuronal or nerve tracts. In other words, they function as a communicational bridge, intercommunicating sensory neurons with motor neurons. Like motor cells, interneurons are only found in the central nervous system. In contrast to the peripheral nervous system, all CNS neurons appear to be interneurons, as they are in communication with many other neurons. However, the term "interneuron" refers to neurons that have axon and dendritic extensions of local extension and not distant, that is, short.

7 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

3 years ago

Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ

bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c, s= 400 m

r= ?

we have to solve for the radius

we know that circumference

$c= 2\pi r$

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular displcament

∅= 400/63.63

∅= 6.28 radians

8 0

3 years ago

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

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4 years ago