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1 year ago

Tyler drives 50km north. Tyler then drives back 30km south. What distance did he cover? What was his displacement?

Physics

1 answer:

Luda [366]1 year ago

7 0

Answer:

he covered 80km his displacement was 20km

Explanation:

displacement is the distance from the starting point so in this case its 20 (50-30) and total distance covered is how many kilometers he drove in total

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Which of the following exists around every object that has mass?

ICE Princess25 [194]

Answer:

A is the correct answer.

Explanation:

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2 years ago

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A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

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2 years ago

Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate

Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

$E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J$

Time required to burn the fat

$t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s$

Distance traveled by the bird

$s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m$

The bird will fly 455165.278 m

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2 years ago

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A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t

LenKa [72]

Answer:

a) W = - 318.26 J, b) W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

sin θ = Wₓ / W

Wₓ = W sin 60

cos θ = Wy / W

Wy = W cos 60

X axis

How the body is going at constant speed

fr - Wₓ = 0

fr = mg sin 60

fr = 15 9.8 sin 60

fr = 127.31 N

Y Axis

N - Wy = 0

N = mg cos 60

N = 15 9.8 cos 60

N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

W = mg L cos θ

Where the angles are between the weight and the displacement is

θ = 60 + 90 = 150

W = 15 9.8 2.50 cos 150

W = - 318.26 J

b) The work of the normal force

From Newton's equations

N = Wy = W cos 60

N = mg cos 60

W = N L cos 90

W = 0

c) The work of the friction force

W = fr L cos 0

W = 127.31 2.50

W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

W = F L cos 0

W = 127.31 2.50

W = 318.275 J

e) the net force

F ’= fr - Wx = 0

W = F ’L cos 0

W = 0

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2 years ago

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
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2 years ago

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