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erastova [34]
3 years ago
6

Two cars are traveling along a straight line in the same direction, the lead car at 24.7m/s and the other car at 29.9m/s. at the

moment the cars are 38.1m apart, the lead driver applies the brakes, causing her car to have an acceleration of -1.96m/s2. how long does it take for the lead car to stop? (in s)
Physics
1 answer:
sattari [20]3 years ago
7 0

Answer: 12.6 s

We would use the equation of motion:

v-u=at

where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

It is given that the lead car has an initial velocity, u=24.7\hspace{1mm} m/s. It decelerates with a=-1.96 \hspace{1mm} m/s^{2} to stop (v=0). we need to calculate the time taken by the lead car to stop.

\Rightarrow t=\frac{v-u}{a}\\ \Rightarrow t=\frac{0-24.7}{-1.96}=12.6\hspace{1mm} s

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Answer:

5 remainder 5

Explanation:

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2 years ago
A 2.2 kg object is whirled in a vertical circle whose radius is 1.0 m. If the time of one revolution is 0.97 s,
strojnjashka [21]

Answer:

The tension in the string at the top  =  71 N

The tension in the string at the top  =  1.1 \times 10^{-2} N

Explanation:

a) at the top  

F_{cent} = F_g +F_T ----------------------(1)

Where,

F_{cent} is the centripetal force

F_g is the gravitational force

F_T  force due to tension

From (1)

F_T = F_{cent} - F_g

F_T = \frac{ 4\pi^2 Rm}{T^2} -mg

where

R is the radius

m is the mass

T is the time taken for one revolution

g is the acceleration due to gravity

On Substituting the values

F_T = \frac{4 (3.14)^2 (1.0)}{(0.97)^2} -(2.2 \times9.8)

F_T= 70.7259N

F_T =71N

b) at the bottom

On Substituting the values

F_{cent} = -F_g +F_T

F_T = F_{cent}- F_g

F_T = \frac{ \pi^2 Rm}{T^2} +mg

F_T = \frac{4 (3.14)^2 (1.0)}{(0.97)^2} +(2.2 \times9.8)

F_T  = 113.8899N

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8 0
3 years ago
Three beads are placed along a thin rod. The first bead, of mass m1 = 28 g, is placed a distance d1 = 1.5 cm from the left end o
Vladimir79 [104]

Answer:

Part a)

Center of mass with respect to the left end is given as

r_{cm} = 5.63 cm

Part b)

Center of mass with respect to middle bead is

r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}

Part c)

Center of mass with respect to middle bead is

r_{cm} = 1.63 cm

Explanation:

Part a)

As we know that the center of mass of the system of mass is given by the formula

r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}

here we have

m_1 = 28 g

m_2 = 11 g

m_3 = 45 g

r_1 = 1.5 cm

r_2 = 1.5 + 2.5 = 4 cm

r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm

Now we have

r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}

r_{cm} = 5.63 cm

Part b)

As we know that the center of mass of the system of mass is given by the formula

r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}

here we have

m_1 = 28 g

m_2 = 11 g

m_3 = 45 g

r_1 = -d_2 = -2.5cm

r_2 = 0

r_3 = d_3 = 4.6 cm

r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}

Part c)

Now plug in the values in above formula

r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}

r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}

r_{cm} = 1.63 cm

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