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Art [367]
3 years ago
11

I'm giving out a lot of points for whoever helps me and answers correctly! 3 questions to answer! Please answer them ALL!!! :D

Physics
2 answers:
lions [1.4K]3 years ago
5 0

1-a 2-b 3-b    

srry had to write more, i go to this school














                                                                                                                 

Lemur [1.5K]3 years ago
3 0

1. A) A ball sits motionless on the ground

Newton's First Law of inertia states that:

"When the forces acting on an object are balanced, the object stays at rest (if it was at rest) or continues its motion with constant velocity (if it was in motion)"

Among the options given, we see that the option

A) A ball sits motionless on the ground

is an example of this law: in fact, the ball sits at rest on the ground, the forces acting on it are balanced, so the ball stays at rest.

2. B) The earth pulls more on the shuttle

The gravitational attraction exerted by a planet on the shuttle is given by

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the shuttle

r is the distance between the shuttle and the planet

In this problem, we are comparing the gravitational pulls exerted by the Earth and the Moon on the shuttle. The distance, r, is the same (because the shuttle is half-way), the mass of the shuttle m is the same, and G is the same. However, the mass of the Earth is greater than the mass of the Moon, so in the formula M is greater for the Earth: therefore, the Earth pulls more on the shuttle.

3. B) The velocity of the car reduced from 50 km/h to 35 km/h over one minute

The acceleration of the car is defined as the ratio between the change in velocity and the time taken:

a=\frac{v_f-v_i}{\Delta t}

where vf is the final velocity, vi is the initial velocity, and \Delta t is the time taken.

In order to have a negative acceleration, the numerator must be negative: this means that the initial velocity, vi, must be greater than the final velocity, vf, and this situation occurs in option B), since the initial velocity is 50 km/h while the final velocity is 35 km/h.

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What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
jek_recluse [69]

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = 1 \times 10-12 W/m^2

Now

Intensity level ( or Loudness)is

L = log10 \frac{I}{10}

L = log10 \frac{10}{1\times 10^{-12}}

L = log10 \times 1013

= 13 \times 1 ( log10(10) = 1)

Therefore  

L = 13 bel

And as we know that

1 bel = 10 decibels

So,

The  Sound intensity level is

L = 130 decibels

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3 years ago
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tigry1 [53]
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During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25
Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

dQ=0

dU=-dW

We need to calculate the number of moles and specific heat

Using formula of heat

dU=nC_{v}dT

nC_{v}=\dfrac{dU}{dT}

Put the value into the formula

nC_{v}=\dfrac{-100}{5}

nC_{v}=-20\ J/K

We need to calculate the heat

Using formula of heat

dQ=nC_{v}(dT_{1})+dW_{1}

Put the value into the formula

dQ=-20\times5+25

dQ=-75\ J

We need to calculate the heat capacity for the second process

Using formula of heat

dQ=nC_{v}(dT_{1})

Put the value into the formula

-75=nC_{v}\times(-5)

nC_{v}=\dfrac{-75}{-5}

nC_{v}=15\ J/K

Hence, The heat capacity for the second process is 15 J/K.

5 0
3 years ago
When the switch is closed, what happens to the current in the circuit? (The intrinsic resistance of the inductor is zero.)
Hitman42 [59]
The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>When the switch is closed, the current instantaneously becomes 1.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
6 0
3 years ago
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the
Serjik [45]
The elastic potential energy of a spring is given by
U= \frac{1}{2}kx^2
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
W=-\Delta U =  \frac{1}{2}kx_i^2 -  \frac{1}{2}kx_f^2

In our problem, initially the spring is uncompressed, so x_i=0. Therefore, the work done by the spring when it is compressed until x_f is
W=- \frac{1}{2}kx_f^2
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
8 0
3 years ago
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