On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer 1
Answer:

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J


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What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that v = f * d where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

How long is a bus route across a small town

Answers

Answer:

4 kilometers

Explanation:

Final answer:

The length of a bus route in a small town can greatly depend on the specifics of the town and the route. It can range from a couple miles in a very small town to 20 miles or more for larger towns.

Explanation:

The length of a bus route across a small town can vary greatly depending on the size of the town and the specifics of the bus route. In a very small town, the bus route might only be a mile or two long. For larger towns, it could easily be 10-20 miles, or more. If you know the specifics of the route (streets it travels along, the number of stops, etc.), you could use a tool like Go_gle Maps to calculate an approximate distance.

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The gravitational force law, deduced by Newton in the 1660's, is remarkably similar to Coulomb's law. Recall that the universal law of gravitation states that the magnitude of the gravitational force between two masses M1 and M2 separated by a distance R is given by the following equation:________. F = G (M1 x M2) / R2
G = 6.67 x 10-11 Nm2/kg2
a. Calculate the value of the gravitational force between an electron (mass = 9.11 x 10-31 kg) and a proton (mass is 1836 times greater than the mass of an electron) if the two particles are separated by 3.602 nanometers. (1 nanometer or 1 nm = 1 x 10-9 m)
F= ______ N
b. The force created in the above question is:
1. repulsive
2. attractive

Answers

Answer:

a.F=7.83* 10^(-51) N

b.Attractive

Explanation:

We are given that

F=(GM_1M_2)/(R^2)

G=6.67* 10^(-11) Nm^2/kg^2

Mass of an electron,M_1=9.11* 10^(-31) kg

Mass of proton,M_2=1836* 9.11* 10^(-31) kg

Distance between electron and proton,R=3.602nm=3.602* 10^(-9) m

1nm=10^(-9) m

a.Substitute the values then  we get

F=(6.67* 10^(-11)* 9.11* 10^(-31)* 1836* 9.11* 10^(-31))/((3.602* 10^(-9))^2)

F=7.83* 10^(-51) N

b.We know that like charges repel to each other and unlike charges attract to each other.

Proton and electron are unlike charges therefore, the force between proton and electron is attractive.

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.

Answers

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= (1)/(2)at^2

h= (1)/(2)4.71*4.96^2

h= 58.012 m

h≅ 58 m

An AC voltage source has an output of ∆V = 160.0 sin(495t) Volts. Calculate the RMS voltage. Tries 0/20 What is the frequency of the source? Tries 0/20 Calculate the voltage at time t = 1/106 s. Tries 0/20 Calculate the maximum current in the circuit when the generator is connected to an R = 53.8 Ω resistor.

Answers

Answer:

RMS voltage is 113.1370 V

frequency is 780.685 Hz

voltage is −158.66942 V

maximum current is  2.9739 A

Explanation:

Given data

∆V = 160.0 sin(495t) Volts

so Vmax = 160

and angular frequency = 495

time t = 1/106 s

resistor R = 53.8 Ω

to find out

RMS voltage and frequency of the source and  voltage  and maximum current

solution

we know voltage equation = Vmax sin ωt

here Vmax is 160 as given equation in question

so RMS will be Vmax / √2

RMS voltage = 160/ √2

RMS voltage is 113.1370 V

and frequency = angular frequency / 2π

so frequency = 497 / 2π

frequency is 780.685 Hz

voltage at time (1/106) s

V(t) = 160.0 sin(495/ 108)

voltage = −158.66942 V

so current from ohm law at resistor R 53.8 Ω

maximum current = voltage max / resistor

maximum current =  160 / 53.8

maximum current =  2.9739 A

Final answer:

The root-mean-square voltage of the AC source is 113.14 V, its frequency is 78.75 Hz, and the voltage at time t = 1/106 s is approximately 150.4 V. The current at this peak voltage, when connected to a resistor of 53.8 Ω, is approximately 2.97 A.

Explanation:

The output of an AC voltage source can be represented by the equation V = V₀ sin ωt, where V₀ is the peak voltage, ω is the angular frequency, and t is the time. In this case, V₀ = 160 V and ω = 495 (1/s). The root-mean-square voltage (Vrms), which is commonly used to express AC voltage, can be calculated from the peak voltage using the formula Vrms = V₀/√2 which gives approximately 113.14 V.

The frequency of the source is related to the angular frequency by the equation f = ω/2π, which gives a frequency of approximately 78.75 Hz. To find the voltage at a specific time t = 1/106 s, we substitute these values into the initial equation resulting in V = V₀ sin ωt = approximately 150.4 V.

Finally, the resistance R = 53.8 Ω allows us to calculate the maximum current in the circuit given by I = V/R. The maximum current occurs at the peak voltage, so I(max) = V₀/R = approximately 2.97 A.

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As shown in the figure below, Greta walks 30m toward her truck. She notices she forgot hercoffee and returns back to the house. Her total travel time is 240 seconds.
30 m
30 in
What is Greta's average velocity over the 240s period?
m/s
What is Greta's average speed over the 240s period?
m/s

Answers

Average velocity: 0 m/s. Average speed: 0.25 m/s. Greta returns to her starting point, so her displacement is 0m.

Greta's average velocity is 0 m/s because she ends up at the same point where she started. Her displacement is 0 meters, and since velocity is displacement divided by time, her average velocity is 0 / 240 = 0 m/s.

Her average speed, on the other hand, is calculated using the formula: Average Speed = TotalDistance / Total Time.

Initially, Greta walks 30 meters away from her truck, and then she returns 30 meters back to her starting point. So, the total distance she covers is 30 + 30 = 60 meters. Her total travel time is 240 seconds.

AverageSpeed = 60 meters / 240 seconds = 0.25 m/s.

In summary, Greta's average velocity is 0 m/s because her net displacement is 0 meters. Her average speed is 0.25 m/s because she covers a total distance of 60 meters in 240 seconds.

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Answer:

0 | for Velocity

.25 | for speed

Explanation: