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1 year ago

A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860

Physics

1 answer:

Charra [1.4K]1 year ago

3 0

Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

Momentum=0.162 x 860

Momentum=139.32kgm/s

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A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

DedPeter [7]

Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

The knowable variables are

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are equally spaced, the distance between each sign is $\frac{25}{3} m$

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

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1 year ago

If a force of 26 N is exerted on two balls, one with a mass of 0.52 kg and the other with a mass of 0.78 kg, the ball with the m

gogolik [260]

False is the correct answer

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1 year ago

A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa

gavmur [86]

Answer : The correct option is, (d) $535^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of copper = $0.10cal/g^oC$

$c_2$ = specific heat of water = $1.00cal/g^oC$

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

$T_f$ = final temperature of mixture = $35^oC$

$T_1$ = initial temperature of copper = ?

$T_2$ = initial temperature of water = $15^oC$

Now put all the given values in the above formula, we get:

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

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Answer:

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Answer:

In hot gases , the atoms keeps colliding with each other and sometimes the energy liberated during collision takes the electron to a higher level,thus, .The object is a cloud of hot gas and finally the electron returns back emitting photon

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