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S_A_V [24]
3 years ago
6

A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

riction force of 15N. The block then slides up a frictionless ramp at 30 degrees reaching a 1.5m maximum height off the ground. What was the force constant of the spring
Physics
1 answer:
Setler [38]3 years ago
6 0

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

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What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz
deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

7 0
3 years ago
A tennis ball is dropped from a height of 3 m and bounces back to a height of
julsineya [31]

Answer:

To decide where the balls land, we need to determine how long the balls are in the air. Both balls will take 2 seconds to hit the ground.

Explanation:

1) Time played forward : gravity & drag forces are in opposite directions so it takes a longer time to reach the ground. 2) Time played backward : gravity & drag forces are in the same direction so it takes a shorter time to reach the ground.

5 0
2 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
3 years ago
You watch a distant lady driving nails into her front porch at a regular rate of 1 stroke per second. You hear the sound of the
Igoryamba

Answer:

λ = 1360 m

Explanation:

Given data:

frequency of driving nails is given as 1 stroke per second mean at every 0.25 sec she hit the nails

speed of sound is given as 340 m/s

we know that the wave equation is given as

Speed = frequency × wavelength,

v = f × λ

where,

v = speed in meters/second (m/s)

f = frequency in Hertz (Hz)

substituing value to get wavelength  of her driving nails

340 m/s = (1Hz)\times  \lambda

\lambda = \frac{340}{0.25}

λ = 1360 m

4 0
3 years ago
Read 2 more answers
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
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