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4 years ago

10

If one bullet is dropped from a certain height and another is fired from a gun horizontally from the same height which one will

hit the ground first or will they hit at the same time?

1 answer:

Zinaida [17]4 years ago

7 0

If the ground is flat, and both bullets are released at the same time from the same height, then they both hit the ground at the same time.

The horizontal motion of the one from the gun has no effect on its vertical motion.

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Explain two ways of magnetising an object

Nina [5.8K]

Answer:

There are two methods generally used to magnetize permanent magnets: static magnetization and pulse magnetization.

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3 years ago

Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 7

mafiozo [28]

Answer:

1.5 × 10³⁶ light-years

Explanation:

A certain square region in interstellar space has an area of approximately 2.4 × 10⁷² (light-years)². The area of a square can be calculated using the following expression.

A = l²

where,

A is the area of the square

l is the side of the square

l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

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3 years ago

In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is

tresset_1 [31]

Answer:

The expression for the initial speed of the fired projectile is:

$\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})$

And the initial speed ratio for the 9.0mm/44-caliber bullet is 1.773.

Explanation:

For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.

First Phase: Impact

In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:

$P_0=P_f\\v_0m=v_i(m+M)\\v_0=v_i\frac{m+M}{m}$ (1)

Second Phase: pendular movement

After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:

$Em_i=Em_f\\\frac{1}{2}mv^2_i=mgH\\v_i=[2gH]^\frac{1}{2}$ (2)

The height of the pendulum respect L and θ is:

$H=L(1-cos(\theta))$ (3)

Using equations (1),(2) and (3):

$\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})$ (4)

The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):

$\displaystyle \frac{v_{9mm}}{v_{44}} =\frac{(M+m_{9mm})m_{44}}{(M+m_{44})m_{9}}(\frac{1-cos(\theta_{9mm})}{1-cos(\theta_{44})} )^{\frac{1}{2}}=1.773$

3 0

3 years ago

A vertical piston-cylinder device initially contains 0.1 m^3 of air at 400 K and 100 kPa. At this initial condition, the piston

jenyasd209 [6]

Answer:

$Q=-38.15kJ$

Explanation:

From the question we are told that

Piston-cylinder initial Volume of air $v_1=0.1 m^3$

Piston-cylinder initial temperature $T_1=400k$

Piston-cylinder initial pressure $P_1= 100kpa$

Supply line temperature $T_s=400k$

Supply line pressure $P_s= 500kpa$

Valve final pressure $P_v=500kpa$

Piston movement pressure $P_m=200kpa$

Piston-cylinder final Volume of air $v_2=0.2 m^3$

Piston-cylinder final temperature $T_2=440k$

Piston-cylinder final pressure $P_2= 500kpa$

Generally the equation for conservation of mass is mathematically given by

$Q=m_2 \mu_2-m_1 \mu_1 +W-(m_2-m_1)h$

where

Initial moment

$m_1=\frac{p_1 V_1}{RT_1}$

$m_1=\frac{100*0.1}{0.287*400}$

$m_1=8.7*10^-^2kg$

Final moment

$m_2=\frac{p_2 V_2}{RT_2}$

$m_1=\frac{500*0.3}{0.287*440}$

$m_1=79*10^{-2}kg$

Work done by Piston movement pressure

$W=Pd$

$W=P(v_2-v_1)$

$W=200(0.2-0.1))$

$W=20000J$

Heat function

$h=cT_1$

$h=1.005(400)$

$h=402$

Therefore

$Q=(0.792*0.718(440)-0.0871*0.718(400)+20-(0.792-0.087)*402))$

$Q=-38.15kJ$

It is given mathematically that the system lost or dissipated Heat of

$Q=-38.15kJ$

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3 years ago

two children of masses m1 = m2 = 20 kg , ride on the perimeter of a small merry-go- round . the merry -go-round us a disk of mas

MrMuchimi

Answer:

c

Explanation:

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3 years ago