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4 years ago

A Galapagos turtle travels 7 cm/s. How long would it take the turtle to travel 90 m (that's

Physics

1 answer:

Marta_Voda [28]4 years ago

8 0

Explanation:

first convert 90m into cm

9000cm × 7cm = x

x= 63000 seconds

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4 years ago

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4 years ago

An 18-kg bicycle carrying a 62-kg girl is traveling at a speed of 7 m/s. What is the kinetic energy of the girl

Dafna11 [192]

Answer:

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3 years ago

A barge ﬂoating in fresh water (p = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H

Karolina [17]

Answer:

a) $\Delta m = 330000\,kg$ , b) $h = 1.268\,m$

Explanation:

a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:

$\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V$

$\Delta m = \rho_{w}\cdot \Delta V$

$\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)$

$\Delta m = 330000\,kg$

b) The submersion height is found by using the equation derived previously:

$\Delta m = \rho_{w}\cdot \Delta V$

$450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)$

The final submersion height is:

$h = 1.268\,m$

4 0

3 years ago

Read 2 more answers

A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$

Finally, the total work done on the moving particle can be calculated.

$W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J$

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3 years ago

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