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4 years ago

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If a ball is given a push so that it has an initial velocity of 4 m/s down a certain inclined plane, then the distance it has ro

lled after t seconds is given by the following equation. s(t)=4 t+5 t^2 (a) Find the velocity after 2 seconds. 28 Incorrect: Your answer is incorrect. m/s (b) How long does it take for the velocity to reach 45 m/s?

1 answer:

nata0808 [166]4 years ago

5 0

Answer:

a. 24m/s

b.4.1 s

Explanation:

We are given that

Initial velocity of ball= $v_0=4 m/s$

$s(t)=4t+5t^2$

a.We have to find the velocity after 2 s.

Differentiate w.r.t t

$v=\frac{ds}{dt}=4+10t$

Substitute t=2

$v(2)=4+10(2)=24m/s$

b. $v(t)=4+10t$

v(t)=45m/s

Substitute the value

$45=4+10t$

$10t=45-4=41$

$t=\frac{41}{10}=4.1 s$

Hence, the it takes 4.1 s for the velocity to reach 45 m/s.

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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.

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

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$m_B=250\ kg$

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$m_D=100\ kg$

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

$P_A=(0,50)$

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

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$F_{BC}=66.406G\ N$

<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>

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$F_x=66.406G-15.625G$

$F_x=50.781G\ N$ 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>

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

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

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<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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