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4 years ago

13

How long (distance and time) would it take a vehicle to accelerate from a stop to 203 mph with an average acceleration of 2.93ft

/s²

1 answer:

butalik [34]4 years ago

6 0

Answer:

t = 1.68 min

S = 2.82 miles

Explanation:

Given,

The initial velocity of the vehicle, u = 0

The final velocity of the vehicle, v = 203 mph

The average acceleration of the vehicle, a = 2.93 ft/s²

= 7191.82 miles/h²

Using the first equation of motion

v = u + at

t = (v - u) / a

= (203 - 0) / 7191.82

= 0.028 h

= 100.8 s

Thus, the time taken by the vehicle to reach the final velocity is, t = 100.8 s

Using the third equations of motion

S = ut + 1/2 at²

Substituting the values

S = 0.5 x 7191.82 x (0.028)²

= 2.82 miles

Hence, the distance traveled by the vehicle, S = 2.82 miles

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A 2.35-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 22.0 N.

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How does the mass of an astronaut change when she travels from earth to the moon? how does her weight change?

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3 years ago

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Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm)

rusak2 [61]

Answer:

$F_2=2.43\times10^{-19}\ N$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

$m_A=200\ kg$

$m_B=250\ kg$

$m_C=1700\ kg$

$m_D=100\ kg$

Now their respective coordinate positions (in cm) are as given below:

$P_A=(0,50)$

$P_B=(0,0)$

$P_C=(-80,0)$

$P_D=(40,0)$

<u>Now force on B due to A:</u>

$F_{BA}=G\times \frac{200\times 250}{50^2}$

$F_{BA}=20G\ N$

<u>Now force on B due to C:</u>

$F_{BC}=G\times \frac{1700\times 250}{80^2}$

$F_{BC}=66.406G\ N$

<u>Now force on B due to D:</u>

$F_{BD}=G\times \frac{100\times 250}{40^2}$

$F_{BD}=15.625G\ N$

<em>We observe that the forces due to masses C&D act opposite in direction.</em>

<u>So, the net force in the x-direction:</u>

$F_x=F_{BC}-F_{BD}$

$F_x=66.406G-15.625G$

$F_x=50.781G\ N$ in the positive x-direction

<em>We have only one force in y-direction due to mass A.</em>

So,

$F_y=20G\ N$ in the positive y-direction.

<u>Now the net force:</u>

$F_2=\sqrt{F_y^2+F_x^2}$

$F_2=\sqrt{(20G)^2+(50.781G)^2}$

$F_2=54.5776G^2\ N$

$F_2=2.43\times10^{-19}\ N$

<u>Now the direction of this force with respect to x-axis:</u>

$tan\ \theta=\frac{F_y}{F_x}$

$tan\ \theta=\frac{20G}{50.781G}$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

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