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2 years ago

10

A dart in a toy dart gun is pressed 4 cm against a spring. if it has 1.5 J of stored potential energy what is the spring constan

t?

1 answer:

Sphinxa [80]2 years ago

8 0

U = 0.5k(x^2)
1.5 J = 0.5k(0.04m^2)
k = 1875 N/m

Check my math i might be wrong

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14 What is the weight of each of

Gwar [14]

Explanation:

On earth,

g=10 m/s^2

When M=2 kg,

weight on earth=2×10

=20N

When M=3.5 kg,

weight on Earth=3.5×10

=35N

When M=5kg,

weight on Earth=5.25×10

=52.5N

5 0

2 years ago

A 250 mL sample of nitrogen(N2) has a pressure of 745 mm Hg at 30 C.What is the mass of nitrogen?

Lera25 [3.4K]

We assume that nitrogen gas sample here is an ideal gas. So that, we can use the ideal gas equation which is expressed as follows:

PV = nRT

First, we calculate the number of moles from the equation above. Then, use molar mass to calculate mass.

n = PV/RT
n = (745/760) (0.250) / 0.08206 (30+273.15)
n = 0.01 mol

m = 0.01 ( 28) = 0.28 g

3 0

3 years ago

The maximum electric field strength in air is 3.0 Mv/m . Stronger electric fields ionize the air and create a spark. What is the

adoni [48]

Answer:

The value is $P_{max} = 2.11 * 10^{6} \ W$

Explanation:

From the question we are told that

The maximized electric field strength is $E = 3.0 \ MV/m = 3.0 *10^{6} \ V/m$

The diameter is $d = 1.5\ cm = 0.015 \ m$

Generally the cross -sectional area is mathematically represented as

=> $A = \frac{\pi * d^2 }{4}$

=> $A = \frac{3.142 * 0.015^2 }{4}$

=> $A = 0.0001767 \ m^2$

Generally the maximum power is mathematically represented as

$P_{max} = \frac{E_{max}^2}{ 2 * \mu_o * c } * A$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

$\mu_o$ is the permeability of free space with value $\mu_o = 4 \pi *10^{-7} \ H/m$

$P_{max} = \frac{ (3.0 *10^{6})^2}{ 2 * (4 \pi *10^{-7}) * 3.0 *10^{8} } * 0.0001767$

=> $P_{max} = 2.11 * 10^{6} \ W$

4 0

2 years ago

A moving object has a kinetic energy of 150J and a momentum of 20.3kgxm/s find the speed of the object in m/s

Irina-Kira [14]

Answer:

$v = 14.78m/s$

Explanation:

Given

$KE = 150J$

$p = 20.3kgm/s$

Required

Determine the object's speed

Kinetic Energy is calculated as:

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

Make m the subject

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

Momentum is calculated as:

$p = mv$

Make m the subject

$m = \frac{p}{v}$

So, we have:

$m = \frac{p}{v}$ and $m = \frac{2KE}{v^2}$

Equate both expressions: $m = m$

$\frac{2KE}{v^2} = \frac{p}{v}$

Multiply both sides by v

$v * \frac{2KE}{v^2} = \frac{p}{v}*v$

$\frac{2KE}{v} = p$

Make v the subject

$v = \frac{2KE}{p}$

Substitute $KE = 150J$ and $p = 20.3kgm/s$

$v = \frac{2 * 150}{20.3}$

$v = \frac{300}{20.3}$

$v = 14.78m/s$

5 0

2 years ago

On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and expl

fredd [130]

Answer:

156.67 m/s

0.45676 times the speed of sound

Explanation:

Distance from the ground = 23.5 km = 23500 m

Time taken by the blast waves to reach the ground = $2\ minutes\ 30\ seconds=2\times 60+30=150\ s$

Spedd of the wave would be

$Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s$

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

$\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v$

The blast wave is 0.45676 times the speed of sound

7 0

3 years ago