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dsp73
3 years ago
8

What is 3,278,000 as scientific notation?

Physics
1 answer:
jonny [76]3 years ago
4 0
3.278*10^6 I think. Sorry if it’s wrong.
You might be interested in
1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.
Masja [62]

Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

K=\frac{1}{2}mv^2

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

K=\frac{1}{2}(0.40)(16)^2=51.2  J

The potential energy of the ball is given by

U=mgh

where

m = 0.40 kg

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

h = 22 m is the heigth of the cliff

Substituting,

U=(0.40)(9.8)(22)=86.2 J

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

E=K+U=51.2 + 86.2=137.4 J

4 0
3 years ago
Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees

ELECTRIC FORCE (F)

F = \frac{KQq}{d^{2} }

Where K = 9 x 10^{9} Nm^{2}/C^{2}

The distance between q_{1} and q_{3} can be calculated by using Pythagoras theorem.

d = \sqrt{33^{2} + 33^{2}  }

d = 46.7 cm = 0.467 m

For force F_{1}, substitute all the parameters into the formula above

F_{1} = (9 x 10^{9} x 3 x 1)/0.467^{2}

F_{1} = 2.7 x 10^{10}/0.218

F_{1} = 1.24 x 10^{11} N

For force F_{4}, substitute all the parameters into the formula above

F_{4} = (9 x 10^{9} x 3 x 4)/0.33^{2}

F_{4} = 1.08 x 10^{11}/0.1089

F_{4} = 9.92 x 10^{11} N

For force F_{2}, substitute all the parameters into the formula above

F_{2} = (9 x 10^{9} x 3 x 2)/0.33^{2}

F_{2} = 5.4 x 10^{10}/0.1089

F_{2} = 4.96 x 10^{11} N

Summation of forces on Y component will be

F_{y} = F_{4} - F_{1} Sin 45

F_{y} = 9.92 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{y} = 9.04 x 10^{11} N

Summation of forces on X component will be

F_{x} = F_{2} - F_{1} Cos 45

F_{x} = 4.96 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{x} = 4.08 x 10^{11} N

Net Force = \sqrt{F_{x} ^{2} + F_{y} ^{2}  } }

Net force = \sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2}  }

Net force = 9.9 x 10^{11} N

The direction will be

Tan ∅ = F_{y}/F_{x}

Tan ∅ = 9.04 x 10^{11} / 4.08 x 10^{11}

Tan ∅ = 2.216

∅ = Tan^{-1}(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0
2 years ago
Read 2 more answers
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau
AleksandrR [38]

We know, by conservation of energy :

\dfrac{kx^2}{2}=mgh

Therefore,

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}

Putting given values, we get :

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0
2 years ago
North Dakota Electric Company estimates its demand trend line​ (in millions of kilowatt​ hours) to​ be: D​ = 75.0 ​+ 0.45​Q, whe
Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line​ is

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

We need to calculate the demand forecast for winter

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times101)\times0.80

D=96.36\ millions\ KWH

We need to calculate the demand forecast for spring

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times102)\times1.20

D=145.08\ millions\ KWH

We need to calculate the demand forecast for summer

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times103)\times1.40

D=169.89\ millions KWH

We need to calculate the demand forecast for fall

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times104)\times0.60

D=73.08\ millions KWH

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0
3 years ago
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
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