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torisob [31]
3 years ago
15

You riding in a car traveling at a speed of 73 mi/hr in a. 60 mi/hr zone, when suddenly you spot a police car parked 20 ft ahead

of you. Do you hit the brakes and slow down with an acceleration of -2 ft/sec,for a total time of 3 seconds. Do you get a ticket?
Physics
2 answers:
fenix001 [56]3 years ago
5 0
I think yes i do.. cuz using yr information i get the distance i travelled before stopping become 49.62144 m, using the formula distance travelled/time taken= (final velocity+initial velocity)/2.
I am not sure tht its the right answer but using my calculation I got this so hope there! :)
Nataliya [291]3 years ago
4 0

Answer:

You get a ticket you speed is around 68 mi/hr

Explanation:

V_{f} =V_{i} + a*t

V_{i} = 73 \frac{mi}{hr} *\frac{1 hr}{3600 s} * \frac{1609,34m}{1 mi} = 32.6339 \frac{m}{s}, a=-2 \frac{ft}{s^{2} } * \frac{0.3048m}{1 ft} =-0.6096 \frac{m}{s^{2} }

V_{f} =32.6339\frac{m}{s}  -0.6096 \frac{m}{s^{2} } *3 s

V_{f} = 30.8051\frac{m}{s} *\frac{3600 s}{hr} * \frac{1 mi}{1609.34m} = 68.909 \frac{mi}{hr}

You still in trouble because  the speed is still above the allowed

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Elements from which two groups in the periodic table would most likely combine
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The answer is actually 1 & 17, maybe you meant to put 1&17 as option b
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2 years ago
A 513 g ball strikes a wall at 14.7 m/s and rebounds at 11.3 m/s. The ball is in contact with the wall for 0.038 s. What is the
Natasha_Volkova [10]

consider the velocity of the ball towards the wall as negative and away from the wall as positive.

m = mass of the ball = 513 g = 0.513 kg

v₀ =  initial velocity of the ball towards the wall before collision = - 14.7 m/s

v = final velocity of the ball away from the wall after collision = 11.3 m/s

t = time of contact with the wall = 0.038 sec

F = average force acting on the ball

using impulse-change in momentum equation , average force is given as

F = m (v - v₀)/t

inserting the values

F = (0.513) (11.3 - (- 14.7))/0.038

F = 351 N


5 0
3 years ago
A ball of mass m = 0.1kg is connected to a rope of length L = 1.2 m. The ball is swung around in a vertical circle and ball is m
bonufazy [111]

Answer:

The speed of the ball is approximately 5.94 m/s

The Tension of the string at the bottom is 3.92 N

Explanation:

We need to find the speed of the ball, which is constant due to the fact that we are in a uniform circular motion. Notice as well that the speed of the ball is the magnitude of the tangential velocity "v_t" (vector that changes direction with the position of the ball but doesn't change magnitude in this case).

We analyze first the top position of the circular motion, for which information on the tension of the string is given (see first free body diagram in the attached picture).  We are told that the tension at the top of the movement equals twice the force of gravity on the ball's mass: T - 2*m*g = 1.96 N. And we know that there are two forces acting on the ball in that position (illustrated with the green arrows pointing down): one is the ball's weight due to gravity, and the other is the string's tension. So we can write Newton's second law for this situation:

F_{net}= T_{top}+W\\F_{net}=2\,W+W\\F_{net}=3\,W\\F_{net}=2.94 N\\

Newton's second law tells us that the net force should equal the mass of the ball times its acceleration (F = m * a), and in this motion, the acceleration is the centripetal acceleration. Therefore weuse this equation to solve for the centripetal acceleration of the ball:

m\,a_c=2.94\,N\\a_c=\frac{2.94\,N}{0.1\,kg} \\a_c=29.4\,\frac{m}{s^2}

The centripetal acceleration is defined as the square of the tangential velocity divided the radius of the circular motion. Then we use it to derive the magnitude of the tangential velocity (speed of the ball):

a_c=\frac{v^2}{R} \\29.4\,\frac{m}{s^2} =\frac{v_t^2}{R} \\v_t^2=29.4\,(1.2)\,\frac{m^2}{s^2} \\v_t=5.94\,\frac{m}{s}

So we have found the speed of the ball.

Now we focus our attention to the bottom of the motion, and again use Newton's second law to solve for the string tension (see second free body diagram in the attached picture).

We notice here that the tension and the weight are acting in opposite directions, so we have such into account when finding the net force on the ball, and then solve for the tension knowing the value of the centripetal acceleration (recall that the magnitude of the tangential velocity is the same because of the uniform circular motion).

F_{net}= T_{bot}-W\\m\,a_c=T_{bot}-0.98\,N\\2.94\,N=T_{bot}-0.98\,N\\T_{bot}=(2.94+0.98)N\\T_{bot}=3.92\,N

4 0
2 years ago
A horizontal rod (oriented in the east-west direction) is moved northward at constant velocity through a magnetic field that poi
sergij07 [2.7K]

Answer:

a. The east end of the rod is at higher potential than the west end.

Explanation:

The horizontal rod is oriented in the east-west direction. This means that applying right hand rule, the current will flow from the east to the west. Now, if we assume tat it is a closed loop, we know from polarity of voltage that current usually flows positive to negative terminal within a circuit.

This means the east is at a largely positive terminal while the west is at a largely negative terminal.

Thus, we can say that the east end of the rod is at higher potential than the west end of the rod.

6 0
3 years ago
Two identical blocks are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictio
Radda [10]

Answer:v

Explanation:

Given

Initially spring mass system is moving towards right with velocity v

Now if the spring is released left block moves towards left with a velocity \frac{v}{2}

As there is no external force applied on the system therefore the change in linear momentum is zero i.e. it is conserved

so center of mass continue to move towards the right with velocity v or we can say right block moves towards right with velocity v_{com}+\frac{3v}{2} and left block moves with velocity v_{com}-\frac{3v}{2}

5 0
2 years ago
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