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3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at

5.2 m/s ?

Physics

1 answer:

noname [10]3 years ago

3 0

$Given:\\m_b=100kg\\m_c=1600kg\\v_c=5.2 \frac{m}{s} \\p_b=p_c\\\\Find:\\v_b=?\\\\Solution:\\\\p_b=p_c\\\\p=mv\\\\m_bv_b=m_cv_c\Rightarrow v_b= \frac{m_cv_c}{m_b} \\\\v_b= \frac{1600kg\cdot5.2 \frac{m}{s} }{100kg} =83.2 \frac{m}{s}$

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A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 8.0 m/s2. A 20 kg mas

nadya68 [22]

Answer:

a) 60 N

b) 860 N

Explanation:

Given that,

$m_1$ = 100 kg

$m_2$ = 20 kg

$a_{1}$ = 8.0 $\frac{m}{s^{2}}$

$a_{2}$ = 3.0 $\frac{m}{s^{2}}$

a) By Newton's Law,

∑ $F_{m_2,x} = f_k = m_{2} * a_{2}$

∑ $F_{m_2,y} = F_{N} - m_{2} * g = 0$

Hence,

$f_k = m_2 * a_2 = 20 * 3 = 60 N$

b) By Newton's Law

∑ $F_{m_1,x} = F = m_{1} * a_{1}$

Hence,

$F = m_{1} * a_{1} = 100 * 8 = 800 N$

Net force acting on 100 kg mass,

$F_{net} = F + f_k = 800 + 60 = 860 N$

6 0

3 years ago

If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot

Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=I_{1}+I_{2}$

$I=(m+m_{1})\times r^2+(m+m_{2})\times r^2$

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

$I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2$

$I=117.45\ kg-m^2$

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0

3 years ago

A copper wire has a square cross section 2.0 mm on a side. The wire is 5.0 m long and carries a current of 2.0 A. The density of

kondor19780726 [428]

Answer:

30.22 hours

Explanation:

Given data:

A= l² = (2 x $10^{-3}$ )² = 4 x $10^{-6}$ m²

Length 'L' = 5m

current ' $I$ ' = 2 A

density of free electrons 'n'= 8.5 x $10^{28}$ /m³

Current Density 'J' = $I$ / A

J= 2/4 x $10^{-6}$

J= 5 x $10^{5}$ A/m²

We can determine the time required for an electron to travel the length of the wire by

T= L/ Vd

Where,

L is length and Vd is drift velocity.

Vd can be defined by J/ n|q|

where,

n is the charge-carrier number density

|q| is is the charge carried by each charge carrier =>1.6 x $10^{-19}$ C

T= L/ Vd

Therefore,

T= L . n|q| / J

T= (4 x 8.5 x $10^{28}$ x |1.6 x $10^{-19}$ |)/5 x $10^{5}$

T= 108800 seconds =>1813.33 minutes

Converting minute into hours:

T= 30.22 hours

Thus, time that is required for an electron to travel the length of the wire is 30.22 hours

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3 years ago

Which of the following is true about a planet orbiting a star in uniform circular motion? A. The direction of the velocity vecto

Luda [366]

<span>As it is uniform circular motion therefore speed is constant. Therefore we can rule out option B. Also in circular motion the direction of velocity vector changes therefore velocity can't be constant. Therefore option B is incorrect as well. Also centripetal acceleration is always towards the center so option D is wrong as well. That implies option A is correct.</span>

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4 years ago

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Answer:

Explanation:

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3 years ago