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4 years ago

Which of the following could be classified as a problem that could be answered with a technological design?

Physics

1 answer:

loris [4]4 years ago

4 0

I think the answer is 4) All of the above!! :)

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Which of the following is the best example of a good hypothesis?

Musya8 [376]

Answer:

Which of the following is the best example of a good hypothesis?

B. A cheetah can run faster than a tiger This Is A good Hypothesis Because You Can Test this With a Experiment

xXxAnimexXx

Have a great day!

4 0

3 years ago

Read 2 more answers

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$ . On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

7 0

3 years ago

Two blocks of masses 6 kg and 5.5 kg are

sp2606 [1]

When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.

But linear momentum still has to be conserved. It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.

Momentum = (mass) x (velocity)

After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
(5.5 x 6.8) = 37.4 kg-m/s to the right.

In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.

M x V = (6 x speed) = 37.4 kg-m/s to the left.

Divide each side by 6 : Speed = 37.4 / 6 = 6.2333... m/s left

(That number is (6 and 7/30) m/s .)

5 0

3 years ago

For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equat

zvonat [6]

Answer:

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

Explanation:

The Rydberg equation is given by

$\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

m is the mass of electron

k = 1/4π∈₀

∈₀ = is the permitivity of free space

e is the charge of electron

h is the plank constant

c is the speed of light in vaccum

z is the atomic number = 1

$\frac{1}{\lambda} =R (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹

For Paschen series of H spectrum

$n_f = 3$

$n_i = 5,6,7 ...$

in Paschen series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm$

The brackett series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm$

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

8 0

3 years ago

A runner is jogging in a straight line at a

Fiesta28 [93]

Answer:

Total distance traveled by bird is 10.5 km.

Explanation:

Given that,

Velocity of runner= 8.4 km/hr

Distance = 6.3 km

Velocity of bird = 42 km/hr

Suppose, How far does the bird travel?

We need to calculate the time of bird

Using formula of time

$t=\dfrac{d}{v_{b}}$

Put the value into the formula

$t=\dfrac{6.3}{42}$

$t=0.15\ hr$

We need to calculate the distance runner travels during time period

Using formula of distance

$d=v_{r}\times t$

$d=8.4\times0.15$

$d=1.26\ km$

We need to calculate the distance of runner to FL

Using distance

$d'=6.3-1.26$

$d'=5.04\ km$

We need to calculate the net closing speed between runner and bird

Using speed

$v'=v_{r}+v_{b}$

$v'=8.4+42$

$v'=50.4\ km/hr$

We need to calculate the time for runner and bird to meet

Using formula of time

$t=\dfrac{d}{v'}$

Put the value into the formula

$t=\dfrac{5.04}{50.4}$

$t=0.1\ hr$

We need to calculate the distance covered by runner

Using formula of distance

$d=vt$

Put the value into the formula

$d=8.4\times0.1$

$d=0.84\ km$

We need to calculate the distance covered by bird to runner

Using formula of distance

$d=vt$

Put the value into the formula

$d=42\times0.1$

$d=4.2\ km$

We need to calculate the total distance

Using distance

$D=6.3+4.2$

$D=10.5\ km$

Hence, Total distance traveled by bird is 10.5 km.

4 0

3 years ago