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3 years ago

15

Calculate the potential energy of a 85 kg skier at the top of a 9 m slope. help

1 answer:

velikii [3]3 years ago

4 0

85 x 9.8 m/s2 x 9 = 7,497

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An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8

Aloiza [94]

Answer:

<em>The mass of the apple is 0.172 kg (172 g)</em>

Explanation:

<u>The Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

$m_2v_2-m_2v'=m_1v'-m_1v_1$

Factoring m2 and m1:

$m_2(v_2-v')=m_1(v'-v_1)$

Solving:

$\displaystyle m_2=\frac{m_1(v'-v_1)}{v_2-v'}$

Substituting:

$\displaystyle m_2=\frac{0.043(16.8-84)}{0-16.8}$

$\displaystyle m_2=\frac{-2.8896}{-16.8}$

$\displaystyle m_2=0.172\ kg$

The mass of the apple is 0.172 kg (172 g)

3 0

3 years ago

wolverine [178]

Mechanical energy
I think

5 0

3 years ago

A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B

Novay_Z [31]

Answer:

Wavelength, $\lambda=1.28\times 10^{-14}\ m$

Explanation:

Given that,

Mass of the particle, $m=4.3\times 10^{-28}\ kg$

Acceleration of the particle, $a=2.4\times 10^7\ m/s^2$

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}$

v = a × t

$\lambda=\dfrac{h}{mat}$

$\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}$

$\lambda=1.28\times 10^{-14}\ m$

Hence, this is the required solution.

4 0

3 years ago

Please select the word from the list that best fits the definition

Digiron [165]

Answer:

cellular respiration , note how it says oxygen

5 0

3 years ago

If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial accelerat

Cerrena [4.2K]

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

$F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\ F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2} }{(6.50*10^{-10} )^{2} } \\F=5.46*10^{-10}N$

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

$a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}$

For(b) Acceleration of proton

$a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}$

3 0

3 years ago