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andrew-mc [135]

4 years ago

9

A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan

t of 4.58×105 N/m . When the railroad car comes to a complete stop, the compression of the spring stopper is 32 cm . How fast was the railroad car initially?

1 answer:

In-s [12.5K]4 years ago

8 0

To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

$\text{Initial Energy}=\text{Final Energy}$

$\frac{1}{2} mv^2=\frac{1}{2} kx^2$

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

$mv^2 = kx^2$

$v^2 = \frac{kx^2}{m}$

$v = \sqrt{\frac{kx^2}{m}}$

Our values are,

$m = 5.22*10^4kg$

$k = 4.58*10^5N/m$

$x = 32cm = 0.32m$

Replacing our values we have,

$v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}$

$v = 2.733*10^5m/s$

Therefore the velocity is $2.733*10^5m/s$

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Some kids are playing by rolling a skateboard across level ground beside a 0.75 m high wall. The goal is to drop from the wall v

lys-0071 [83]

Answer:

The speed of the kid and the skateboard together is 0.723 m/s

Explanation:

Let the mass of the kid be $m_{k} = 45 kg$

Let the mass of the skateboard be $m_{s} = 4,5 kg$

The initial speed of the skateboard, $v_{s} = 8 m/s$

The initial horizontal speed of the boy as he jumps on the skateboard, $v_{k} = 0 m/s$

Let the speed of the kid and the skateboard together be v

According to the principle of conservation of momentum:

$m_{k} v_{k} + m_{s} v_{s} = (m_{k} + m_{s}) v$

Substituting the appropriate values into the given equation

$(45*0) + (4.5*8) = (45 + 4.8) v$

36 = 49.8 v

v = 36/49.8

v = 0.723 m/s

7 0

3 years ago

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height

nikdorinn [45]

Answer:

Explanation:

According to Equations of Projectile motion :

$Time\ of\ Flight = \frac{2vsin(x)}{g}$

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

$Maximum Height = \frac{(vsinx)^{2} }{2g}$

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

$Horizontal Range = vcosx * t$

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

4 0

4 years ago

A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) H

SashulF [63]

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

b)

$v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s$

Time taken by the ball to reach the highest point is 2.44 seconds

a)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m$

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s$

The ball reaches the ground 2.57 seconds after reaching the highest point

d)

$v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s$

The ball will hit the ground at 25.2117 m/s

8 0

3 years ago

Why does the band of stability curve upward at high atomic numbers? excess neutrons are required due to the repulsion between th

ella [17]

The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.

Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.

3 0

3 years ago

If a child starts from rest at point A and lands in the water at point B, a horizontal distance L = 2.52 m from the base of the

tamaranim1 [39]

Answer:

The height of the water slide is 0.878 m

Explanation:

Given that,

Distance = 2.52 m

Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$1.80=0+\dfrac{1}{2}\times9.8\times t^2$

$t^2=\dfrac{1.80\times2}{9.8}$

$t=\sqrt{\dfrac{1.80\times2}{9.8}}$

$t=0.606\ sec$

We need to calculate the velocity

Using formula of velocity

$v = \dfrac{d}{t}$

Put the value into the formula

$v=\dfrac{2.52}{0.606}$

$v=4.15\ m/s$

We need to calculate height

Using conservation of energy

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{4.15^2}{2\times9.8}$

$h=0.878\ m$

Hence, The height of the water slide is 0.878 m.

4 0

3 years ago