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3 years ago

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Block A, with a mass of 4.0 kg, is moving with a speed of 3.0 m/s while block B, with a mass of 6.0 kg, is moving in the opposit

e direction with a speed of 5.0 m/s. What is the momentum of the two-block system

1 answer:

ElenaW [278]3 years ago

5 0

Answer:

The momentum of the two-block system is 18 kg-m/s

Explanation:

It is given that,

Mass of block A, $m_A=4\ kg$

Mass of block B, $m_B=6\ kg$

Velocity of block A, $v_A=3\ m/s$

Velocity of block B, $v_B=-5\ m/s$ (it is moving in opposite direction)

We need to find the momentum of two block system. It is given by the product of mass and velocity for both blocks i.e.

$p=m_Av_A+m_Bv_B$

$p=4\ kg\times 3\ m/s+6\ kg\times (-5\ m/s)$

p = -18 kg-m/s

So, the momentum of two block system is 18 kg-m/s. Hence, this is the required solution.

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The length of a wooden rod is 25.5 cm. What is this length in:<br>(a) millimetres?<br>(b) metres?

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Answer:

$(a)255mm \\ (b)0.255m$

Explanation:

$(a)1cm = 10mm \\ 25.5cm = x$

Cross multiply

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$100x = 25.5 \\ \frac{100x}{100} = \frac{25.5}{100} \\ x = 0.255m$

hope this helps

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3 years ago

Read 2 more answers

The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the spe

Snezhnost [94]

Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

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3 years ago

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

NikAS [45]

Answer:

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Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area, $A=0.344 m^2$

Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

Power,P= $(F_d+f)\times v'$

$340=(4.1+4.635) v'=8.735v'$

$v'=\frac{340}{8.735}=38.9m/s$

$v'=87.1 mph$

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