Question 8 of 10It takes a person 22 seconds to swim in a straight line from the south end of
a pool to the north end of the pool, a distance of 28 meters. What is the
swimmer's velocity?
A. 1.3 m/s south
B. 1.3 m/s north
C. 0.8 m/s south
D. 0.8 m/s north

Answers

Answer 1

Answer:

The correct answer is B. 1.3 m/s north

Explanation:

Velocity is a factor that describes how fast or slow the motion of a body occurs and its direction. Moreover, this can be calculated by dividing the total displacement into the time of movement, and the final result is expressed in units such as meters per second followed by the direction, for example, 152 m/s south. The process to calculate the velocity of the swimmer is shown below.

$v = (d)/(t)$

$v = (28 meters)/(22 seconds)$

$v = 1.27 m/s$

This means the velocity of the swimmer is 1.27 m/s, which can be rounded as 1.3 m/s. Additionally, if the direction is considered it would be 1.3 m/s north because the swimmer went from the south of the pool to its north.

Answer 2

Answer:

Answer:

the answer is B

Explanation:

confirmed

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A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Answers

Answer:

her initial frequency is 445 Hz

Explanation:

Given;

initial beat frequency, $F_B$ = 5

observed frequency, F = 440 Hz

let the initial frequency = F₁

F₁ = F ± 5 Hz

F₁ = 440 Hz ± 5 Hz

F₁ = 435 or 445 Hz

This result obtained shows that her initial frequency can either be 435 Hz or 445 Hz

The last beat frequency will be used to determine the actual initial frequency.

F = v/λ

Frequency (F) is inversely proportional to wavelength. That is an increase in length will cause a proportional decrease in frequency.

This shows that the final frequency is smaller than the initial frequency because of the increase in length.

Initial frequency - frequency of tuning fork = 5 beat frequency

Reduced initial frequency - frequency of tuning fork = 3 beat frequency

Initial frequency = 5Hz + 440 Hz = 445 Hz

Final frequency (Reduced initial frequency) = 440 + 3 = 443 Hz

Check: 445 Hz - 440 Hz = 5 Hz

443 Hz - 440 Hz = 3 Hz

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answers

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is $\rm 0.430 \; kg\;m^2$ .

Given :

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

$\rm s = ut + (1)/(2)at^2$

$\rm 1.8 = (1)/(2)* a* (2)^2$

$\rm a = 0.9\; m/sec^2$

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

$\rm \sum F=ma$

$\rm mg-T_b=ma$

$\rm T_b = m(g-a)$

$\rm T_b = 7* (9.8-0.9)$

$\rm T_b = 62.3\;N$

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

$\rm \sum F=ma$

$\rm T_a=ma$

$\rm T_a = 2.1* 0.9$

$\rm T_a = 1.89\;N$

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

$\rm \sum \tau = I\alpha$

$\rm T_br-T_ar = I\alpha$

$\rm I = ((T_b-T_a)r^2)/(a)$

Now, substitute the values of the known terms in the above expression.

$\rm I = ((62.3-1.89)(0.080)^2)/(0.90)$

$\rm I = 0.430 \; kg\;m^2$

For more information, refer to the link given below:

brainly.com/question/2287912

Answer:

(a) 62.3 N

(b) 1.89 N

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B. There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

The Olympias is a reconstruction of a trireme, a type of Greek galley ship used over 2000 years ago. The power P (in kilowatts) needed to propel the Olympias at a desired speed s (in knots) can be modeled by this equation: P = 0.0289s3 A volunteer crew of the Olympias was able to generate a maximum power of about 10.5 kilowatts. What was their greatest speed? Start a New Thread

Answers

Answer:

7.13559 knots

Explanation:

Maximum power = 10.5 kilowatts

$P=0.0289s^3$

where,

P = Power in kilowatts

s = Desired speed in knots

Here, P = 10.5 kW

$10.5=0.0289s^3\n\Rightarrow s^3=(10.5)/(0.0289)\n\Rightarrow s=\left((10.5)/(0.0289)\right)^{(1)/(3)}\n\Rightarrow s=7.13559\ knots$

The greatest speed of the Olympians was 7.13559 knots

Which best describes a reference frame?

Answers

Answer:

C a position from which something is observed

om edu 2021

Explanation:

Answer: A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes.

Suppose you are pushing on a crate across a floor as shown below. Assume the friction force is 47.0 N. How much time will it take for the crate to reach 6.0 m/s if it started from rest? Assume the weight of the crate is 2058 N. (250 N force applied)(Question is no longer a priority but i’d like to know the answer and how it’s found) pls don’t scam i’m serious man i need to know

Answers

Answer:

6.2 seconds

Explanation:

Using Newton's second law, ∑F=ma, we know the net force acting on the object is Force applied-Force of friction. The net force is 203 N. Newton's second law requires the mass of an object, not the weight force, so we will have to calculate the mass. We know that m*g=weight force, in this case, solve for the mass and you will get 210 kg. Now that we have the value of the net force and the mass, we can solve for acceleration. $(F)/(m)=a$ =0.967 m/s^2. Now, since we have the acceleration, initial velocity(0 m/s), and the final velocity (6m/s) we will use these to solve for time using the kinematic equation Vf=Vi + at. Plug in the values we know and solve for time and you will get 6.2 seconds