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3 years ago

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As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0

m in front of him. The ball leaves his foot at a speed of 15.7 m/s and an angle of 31.0 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net.

1 answer:

alekssr [168]3 years ago

6 0

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

$Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s$

The motion on the X axis is a constant velocity motion so:

$t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s$

The whole trajectory of the ball takes 1.48 seconds

We know that:

$Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s$

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

$V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s$

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Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma

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Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ( $C_{HP}$ ) is given by

$C_{HP} = \dfrac{Q_{H}}{W_{in}}$

where, $Q_{H}$ is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature $T_{H}$ and $W_{in}$ is the required input and is given by $W_{in} = Q_{H} - Q_{L}$ , $Q_{L}$ being magnitude of heat transfer between cyclic device and low-temperature $T_{L}$ . Therefore, from above equation we can write,

$&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}$

Given, $T_{L} = 460 K$ and $T_{H} = 540 K$ . So, the minimum work per unit heat transfer is given by

$\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15$

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3 years ago

How much net force is needed to accelerate a 15 kg mass at 2.8 m/s^2

Lera25 [3.4K]

Refer to the 2th Law of Newton

F = m. a

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3 years ago

A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. How much energy is lost? Group of answer choices 0.5

bogdanovich [222]

Answer: A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. Energy loss will be 0.29J

Explanation: To find the correct answer, we have to know more about the Gravitational potential energy.

<h3>What is gravitational potential energy?</h3>

The energy possessed by a body by virtue of its position in gravitational field of earth is called gravitational potential energy.
The gravitational potential energy of a body at a height h with respect to the height h will be,

$U=mgh$

Expression for gravitational potential energy loss will be,

$E=U_i-U_f$

<h3>How to solve the problem?</h3>

The total energy before the ball dropped will be,

$U_i=mgh_i=60*10^-3kg*9.8m/s^2*2m=1.176 J$

The total energy after when the ball rebounds to 1.5m will be,

$U_f=mgh_f=60*10^-3kg*9.8m/s^2*1.5m=0.882J$

The total energy loss will be,

$E=1.176-0.882=0.294J$

Thus, we can conclude that, the energy loss will be,0.294J.

Learn more about the gravitational potential energy here:

brainly.com/question/28044692

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2 years ago

A car is moving at 80km/h accelerates to pass a truck. Five seconds later the car is moving at 100km/h. What is the Acceleration

labwork [276]

Answer/Explanation:

Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. Fortunately, we are given these values. So, we calculate as follows:

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acceleration = (80 mph - 50 mph) ( 1 h / 3600) / 5 s

acceleration = 1.67 x 10^-3 m / s^2

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3 years ago

Read 2 more answers

Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1".The b

Gre4nikov [31]

Answer:

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

See explanation below.

Explanation:

For this case we consider the system as shown on the figure attached.

Since the system is connected the acceleration for both masses are equal, that is $a_{M_1}= a_{M_2} = a$

From the second Law of Newthon we have that the force applied for the mass $M_2$ is $F_{M_2}= M_2 a$ and we know that the force acting on the x axis for the mass 2 is $F_{M_2}= T_2$ so then we have that $T_2= M_2 a$

Now when we consider the system of $M_1 +M_2$ as a whole mass, this system have the same acceleration $a$ and on this case we will see that the only force acting on the entire system would be $T_1$ and then by the second law of Newton we have that:

$F_{M_1 +M_2} = T_1 = (M_1 +M_2) a$

And then if we compare $T_1$ and $T_2$ we see that :

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

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3 years ago