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abruzzese [7]
4 years ago
6

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, startin

g from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?
Physics
1 answer:
a_sh-v [17]4 years ago
8 0

Answer:

Explanation:

Given

Force applied F=12\ N

time taken t_1=2\ s

Displacement s=16\ m

using

s=ut+\frac{1}{2}at^2

u=initial velocity

s=displacement

t=time

16=0+\frac{1}{2}a(2)^2

a=8\ m/s^2

thus mass of body m=\frac{F}{a}=\frac{12}{8}=1.5\ kg

Next it is released from a height of h=10\ m

time taken to reach ground t_2=2.58\ s

using  s=ut+\frac{1}{2}at^2 in vertical direction

here acceleration is due to acceleration due to gravity(g') of the planet

as it at rest so u=0 here

10=0+\frac{1}{2}(g')(2.58)^2

g'=3.004\ m/s^2

Thus acceleration due to gravity on Newtonian is 3\ m/s^2

Weight of tool on Newtonia W'=mg'

W'=1.5\times 3.004=4.506\ N

Weight on Earth W=1.5\times 9.8=14.7\ N

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Answer:

a. Load.

b. Fuse.

c. Source.

d. Wire.

e. Switch.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, fuse, etc.

Matching the given circuit parts to their appropriate functions, we have;

a. Load: an appliance or device that uses electricity source like bulbs, computers, television, radio, etc.

b. Fuse: it is a safety device made from materials that easily melt even before the wires carry too much current.

c. Source: it is where electricity came from like batteries and generators.

d. Wire: it is the pathway of electricity from the resources (source) to the load.

e. Switch: it controls the flow of electricity from the source. It is typically used to turn ON or turn OFF a load.

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3 years ago
An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
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Answer:

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Explanation:

As we know that the power emitted by the source is given as

P = 170 W

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now we know that energy density is given as

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now we have

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intensity is defined as

I = \frac{P}{A}

now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

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I = \frac{P}{4\pi r^2}

I = \frac{170}{4\pi (410)^2}

I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

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3 years ago
What happens to a light if the switch is turned off in the circuit?
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A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
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a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

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Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

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c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

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s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

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Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

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t = 4.11 seconds.

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