I need help 8th grade science test review will give brainest

On Mars a rock falls an unknown vertical distance from a resting position and lands in a crater. If it takes the rock 2.5 second

astra-53 [7]

The Answer To This Question Is B

Hope It Helped

8 0

3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi

tekilochka [14]

Answer:

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg

Explanation:

Hi there!

Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:

PE = EPE

m · g · h = 1/2 k · x²

Where:

m = mass.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression distance

The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:

EPE =1/2 k · x²

EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J

Then, using the equation of gravitational potential energy:

PE = m · g · h = 541.2 J

m = 541.2 J/ g · h

m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)

m = 16.6 kg

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.

6 0

3 years ago

parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet

Rom4ik [11]

Answer:

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

- Te energy density u = 0.037 J /m^3

Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

u = 0.5*ε*E^2

- Where the Electric field strength E between capacitor plates is given by:

E = V / d

Hence,

u = 0.5*ε*(V/d)^2

Where, ε = 8.854 * 10^-12

V^2 = 2*u*d^2 / ε

V = d*sqrt ( 2*u / ε )

Plug in values:

V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )

V = 576 V

4 0

3 years ago