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3 years ago

4. Lead has a density of 11.5g/cm3. A rectangular block of lead measures 7cm ×5cm×2cm.

Physics

1 answer:

MArishka [77]3 years ago

6 0

Answer:

a.) volume = 70cm^3

b.) mass = 805 grams

Explanation:

see the explanation in attachment.

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The magnitude of the tidal force between the International Space Station (ISS) and a nearby astronaut on a spacewalk is approxim

vovikov84 [41]

Answer:

F = 4.47 10⁻⁶ N

Explanation:

The expression they give for the strength of the tide is

F = 2 G m M a / r³

Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg

They ask us to perform the calculation

F = 2 6.67 10⁻¹¹ 135 5.98 10²⁴ 13 / (6.79 10⁶)³

F = 4.47 10⁻⁶ N

This force is directed in the single line at the astronaut's mass centers and the space station

4 0

3 years ago

A lifeguard on a beach observes that waves have a speed of 7 m/s and a distance of 4.6m between wave crests. what is the period

Pepsi [2]

Answer:

0.657 seconds

Explanation:

speed of wave= wavelength / time period

time period= wavelength / speed

= 4.6/7

=0.657 sec

4 0

3 years ago

Suppose there are two identical gas cylinders. One contains the monatomic gas krypton (Kr), and the other contains an equal mass

tensa zangetsu [6.8K]

Answer:

Explanation:

Let equal mass of Ne and Kr be m gm

no of moles of Ne and Kr will be m / 20 and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )

Let the pressure and volume of both the gases be P and V respectively .

The temperature of Ne be T₁ and temperature of Kr be T₂.

For Ne

PV = (m / 20) x R T₁

For Kr

PV = (m / 84) x R T₂

T₁ / T₂ = 84 / 20

We know that

average KE of an atom of mono atomic gas = 3 / 2 x k T

k is boltzmann constant and T is temperature .

KEKr/KENe = T₂ / T₁

= 20 / 84

4 0

3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: