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2 years ago

A 0.03 kg baby sea turtle moves across the sand toward the ocean at 0.3 m/s. What is her kinetic energy in Joules?

Physics

1 answer:

lina2011 [118]2 years ago

4 0

Answer:

.00135 j

Explanation:

K.E. = 1/2 m v^2

= 1/2 * .03 * .3^2 = .00135 j

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Difference between 3.15 m and 2.0 m with the correct number of significant figures?

pentagon [3]

Answer:

1.2

Explanation:

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3 years ago

How far will a car travel in 25 seconds at 10m/ min

Kryger [21]

Time = 25s
speed = 10m/min
= 10m / 60
= (1/6)m/s

distance = speed × time
= 25 × (1/6)
=4.167m

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3 years ago

How can u create a fair test?<br><br> Short answer only:

Crazy boy [7]

Answer:

making sure that you change one factor at a time while keeping all other conditions the same

Explanation:

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2 years ago

Read 2 more answers

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

3 years ago

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

notsponge [240]

Answer:

$Q = \frac{0.068}{E}$

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

$QE = F_x$

$mg = F_y$

since string makes 30 degree angle with the vertical so we will have

$tan\theta = \frac{F_x}{F_y}$

$tan30 = \frac{QE}{mg}$

$Q = \frac{mg}{E}tan30$

$Q = \frac{0.012\times 9.81}{E} tan30$

$Q = \frac{0.068}{E}$

where E = electric field intensity

5 0

3 years ago