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3 years ago

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A boulder with a mass of 50 kg, sits on a cliff that is 50 m high. The boulder is pushed off of the cliff. How much kinetic ener

gy does the boulder have just before it hits the ground?

1 answer:

rosijanka [135]3 years ago

8 0

50 +50 =100 Since it’s sitting on a 50m cliff that’s high with a mass of 50 kg it would be adding because once it goes down it’s adding speed

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a jet plane traveling 1890 km/h pulls out of a dive by moving in an arc of radius 5.20km. what is planes acceleration in g's

Andrew [12]

Answer:

$53 m/s^2$

Explanation:

Unit conversions:

1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s

5.2 km = 5200 m

Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

$a = \frac{v^2}{r}$

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

$a = \frac{525^2}{5200} = 53m/s^2$

7 0

3 years ago

You, a 70 kg person, leap from a 10 m tall building and land feet first on a trampoline. The center of the trampoline where you

Anon25 [30]

Answer:

1373.4 N/m

Explanation:

Hooke's law states that the extension of a spring and force are related by the expression, F=kx where k is spring constant, x is extension of spring and F is the applied force. Making k the subject of the formula then

$k=\frac {F}{x}$

Also, F=gm hence the above formula is modified as

$k=\frac {gm}{x}$

Taking g as 9.81 m/s2 , x as 0.5 m and m as 70 kg then

$k=\frac {9.81\times 70 kg}{0.5m}=1373.4 N/m$

4 0

3 years ago

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with

icang [17]

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of 955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u² - 2.021

u² = 2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

8 0

3 years ago

2225.6 grams =________ ft

AnnyKZ [126]

71606.562467

Is the complete answer

6 0

3 years ago

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago