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2 years ago

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When work is done by an applied force, the object's energy will change. In this Interactive, does the work cause a kinetic energ

y change or a potential energy change?

1 answer:

Alex73 [517]2 years ago

6 0

Answer:

Explanation:

According to the work energy theorem, the work done by all the forces by a body is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

W = Final kinetic energy - initial kinetic energy

So, work causes change in kinetic energy.

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A major benefit of the daguerreotype process is that ________.

tigry1 [53]

Daguerreotype is defined as the first practical photographic process which was introduced in Paris on January 7, 1839. To make the image permanent, Daguerre used salt solution. The result of his introduced process is a finely defined image with surface which is delicate. The major benefit of daguerreotype is if the images are correctly preserved, the pictures could last forever. Aside from this, since it produces superior quality of outline, it is thus suitable for portraitures.

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2 years ago

You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

oksano4ka [1.4K]

As we know that reaction time will be

$t = 0.50 s$

so the distance moved by car in reaction time

$d = vt$

$d = 20 \times 0.50$

$d = 10 m$

now the distance remain after that from intersection point is given by

$d = 110 - 10 = 100 m$

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

$v_f = 0$

$v_i = 20 m/s$

now from kinematics equation we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2(a)100$

$a = \frac{-400}{200} = -2 m/s^2$

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

$v_f - v_i = at$

$0 - 20 = -2 (t)$

$t = 10 s$

total time to stop the car is given as

$T = 10 s + 0.5 s = 10.5 s$

3 0

2 years ago

A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

dlinn [17]

Answer:

a = 4.72 m/s²

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a

f is the friction force

m g sinθ - μ N = ma

m g sinθ - μ m g cos θ = ma

a = g sinθ - μ g cos θ

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²

3 0

2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

2 years ago

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

Strike441 [17]

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer $f_{obs}$ is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

$f_{obs} = f_s (\frac{v_w}{v_w-v_s})$

$f_s$ = Frequency of the source

$v_w$ = Speed of sound

$v_s$ = Speed of source

The velocity of the ambulance is

$v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$v_s = 30.55m/s$

Replacing at the expression to frequency of observer we have,

$f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})$

$f_{obs} = 878Hz$

Therefore the frequency receive by observer is 878Hz

8 0

2 years ago