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2 years ago

Let’s now multiply two numbers in scientific notation using Google.

Physics

2 answers:

Tatiana [17]2 years ago

6 0

Answer

Option 4, 2.33E-11

Step by step solution

Step 1

The first step is to convert the 2 numbers to decimal form as shown below,

4.48E-8= 0.0000000448

5.2E-4=0.00052

Step 2

The next step is to multiply these two numbers by entering them into then in to the google calculator as shown below,

0.0000000448*0.00052=2.329E-11

Step 3

In this step we sound the number up to 2 decimal places. This number to 2 decimal places is 2.33E-11.

olga_2 [115]2 years ago

5 0

We are going to rewrite each number:
(4.48E-8) = 0.0000000448
(5.2E-4) = 0.00052
We observe that when multiplying, the exponent will be on the order of 10 ^ -11
Doing the multiplication we have:
(4.48E-8) * (5.2E-4) = 2.3296E-11
Rewriting:
(4.48E-8) * (5.2E-4) = 2.33E-11
Answer:
2.33E-11

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Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz

Olin [163]

Answer:

The correct answer is option 'd': The frequency decreases and the intensity of the sound decreases.

Explanation:

1) <u>Effect on Frequency </u>

According to Doppler's effect of sound we have

for a source of sound moving away from the observer the relation between the observed and the original frequency is given by

$f_{app}=\frac{c-v_{rec}}{c+v_{s}}\times f_{original}$

where

c = speed of sound in air

$v_{rec}$ is the velocity of observer of sound

$v_{s}$ is the velocity of source of sound

$f_{o}$ is the original frequency of sound

As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.

2) <u>Effect on Intensity:</u>

At a distance 'r' from source emitting a wave of Power 'P' is given by

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

As we see on increasing 'r' intensity of sound decreases.

3 0

2 years ago

How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?

noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

$U=mgh$

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

$U=(50 kg)(9.8 m/s^2)(15 m)=7350 J$

3 0

2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

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

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

2 years ago

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

Gelneren [198K]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Given data

velocity v₀=20 cm/s at time t=3s

velocity vf=0 at time t=8 s

To find

Average Acceleration at time=3s to 8s

Solution

As we know that acceleration is first derivative of velocity with respect to time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

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2 years ago

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

2 years ago