Ask question

Ask question

All categories

2 months ago

15

In pulling two identical carry-on bags through the airport, Mr. Myers and his 13 year old grandson, Vincent, do the same amount

of work. Who has to use a greater force between gates?

1 answer:

serg [3.5K]2 months ago

5 0

Answer:

Mr. Myers and his son exert the same force while pulling the bags through the gates.

Explanation:

The work performed by Mr. Myers when pulling the bags is equivalent to the work his 13-year-old grandson does with the same bag.

Let F₁ be the force Mr. Myers applies and F₂ be the force his grandson applies.

Let d denote the distance across the gate.

Thus, since Work, W = Force, F × Distance, we conclude;

Mr. Myers' work between the gates, W₁ = F₁ × d

His grandson's work in the same situation, W₂ = F₂ × d

Since the amount of work by both Mr. Myers and his grandson is equal, we state;

W₁ = W₂ leading to F₁ × d = F₂ × d, which indicates;

F₁ = F₂, showing that the force each employed while moving through the gates is identical.

You might be interested in

The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

serg [3582]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

6 0

2 months ago

A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa

kicyunya [3294]

A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.

3 0

1 month ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

ValentinkaMS [3465]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is $\frac{4}{3} \pi (\frac{D}{2})^3$

Thus $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

The radius of the disk is $1.15(\frac{ D}{2} )$

Applying angular momentum conservation;

The $M_i$ of the sphere = $\frac{2}{5} m \frac{D}{2}^2$

$M_i$ of the disk = $m*\frac{ \frac{1.15*D}{2}^2 }{2}$

$\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }$

= 0.6

5 0

2 months ago

You are asked to design a spring that will give a 1160-kg satellite a speed of 2.50 m>s relative to an orbiting space shuttle

Maru [3345]

Given the values: m = 1160 kg g = 9.81 m/s² v = 2.5 m/s Unknowns include: k (spring constant) x (spring compression) 1. To address this, use Newton's second law and Hook's law: 2. The total energy in the spring must correspond to the satellite's energy: By combining these two equations

7 0

1 month ago

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia

serg [3582]

Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

$v = \sqrt{2gh}$

We know that

$v = \sqrt{2(9.81)(0.250)}$

$v = 2.21 m/s$

Now applying the conservation of momentum:

$mv = (m + M)v_f$

$m = 0.300 kg$

$M = 0.400 kg$

From this equation, we find:

$0.300 (2.21) = (0.300 + 0.400) v_f$

$v_f = 0.95 m/s$

$0.400 (9.81) = 200 x_1$

When the slice rests on the plate, the new mean position can be expressed as

$x_1 = 0.01962 m$

$(0.300 + 0.400)9.81 = 200 x_2$

We also determine that the speed of SHM is represented as

$x_2 = 0.0343 m$

Here, we derive values from

$v = \omega\sqrt{A^2 - x^2}$

$\omega = \sqrt{\frac{k}{m + M}}$

$\omega = \sqrt{\frac{200}{0.300 + 0.400}}$

$\omega = 16.9 rad/s$

$a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m$

Using the previous formula gives:

$0.95 = 16.9\sqrt{A^2 - 0.0147^2}$

$A = 0.0581 m$

Part b)

The time period for the scale is computed as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{16.9}$

$T = 0.37 s$

8 0

2 months ago