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umka21 [38]
3 years ago
15

A race car is one lap behind the lead race car when the lead car has 44 laps to go in a race. If the speed of the lead car is 55

.9 m/s, what must be the average speed of the second car to catch the lead car just before the end of the race (i.e., right at the finish line)? Assume 1 lap is 1.34 km. Answer in units of m/s
Physics
1 answer:
masha68 [24]3 years ago
3 0

Answer:

Speed of second car will be 57.17 m/sec

Explanation:

We have given lead car travels = 44 laps

1 laps = 1.34 km = 1340 m

So total distance = 1340×44 = 58960 m

Speed of lead car = 55.9 m/sec

We know that time=\frac{distance}{speed}=\frac{58960}{55.9}=1054.74sec

As the second car is 1 lap behind so distance traveled by second car = 45×1340 = 60300 m

So speed of second car will be speed=\frac{distance}{time }=\frac{60300}{1054.74}=57.17m/sec

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Please help! I especially need help with the second question but help with the first one would be most appreciated!
lara31 [8.8K]

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/s^{2}

Explanation:

Given that Team-A pulls with a force , F_{1} = 50N

and Team-B pulls with a force , F_{1} = 45N

∵ F_{1} > F_{2}

The rope will move in the direction of force F_{1}.

∴ Team-A will win.

b) Considering both the teams as one system of total mass , m = 246+253 = 499 kg

Net force on the system , F = F_{1} - F_{2} = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/s^{2}

4 0
3 years ago
The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

5 0
3 years ago
How is thermal energy transferred during convection?
Lerok [7]
Conduction and <span>convection it involves particles.</span>
5 0
3 years ago
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POSSIBLE POINTS: 1.92
gogolik [260]

Answer:

jnfal4u4ryhfsbjls5

Explanation:

duehdakjweyedufkbshegygfr

7 0
3 years ago
Discuss the limitations of using the Doppler shift to determine an object's speed.
pantera1 [17]

Answer and Explanation:

Limitation of Doppler shift :

The Doppler impact is relevant when the speeds of the wellspring of sound and spectator are considerably less than the speed of sound. The movement of both the spectator and the source is along a similar straight line.When movement is not in straight line or velocity is not much less than speed of light then we can not use Doppler shift

This is the limitation of Doppler shift to determine the object distance

3 0
3 years ago
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