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3 years ago

F Penny's piston is wider than Pauly's, who is pushing with greater force?

2 answers:

8 0

<span>A Penny is exerting the greater force.
B Pauly is exerting the greater force.
C They are both using equal force.
Think its A or C</span>

Yuliya22 [10]3 years ago

7 0

It would depend on who is stronger but it is likely that Penny would be creating larger pressure because the surface area is less. If this is part of a question where the pressure for both is the same then Pauly is pushing with greater force, I know this because force/area=pressure and when the surface area is larger the force must be greater as well.

Hope this helps :)

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A body 'A' of mass 1.5kg travelling along the positive X-axis with speed of 4.5m/s collides with another body 'B' of mass 3.2kg,

xz_007 [3.2K]

I already answered this question.
Please refer to this link brainly.com/question/8743596.

6 0

3 years ago

Speed=100m/sec<br> Frequency=10 Hz<br> Wavelength=?

elena-14-01-66 [18.8K]

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 m $s^{-1}$

Frequency = 10 Hz = 10 $s^{-1}$

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

v = fλ

Therefore wavelength λ = v/f

= 100 m $s^{-1}$ / 10 m $s^{-1}$

= 10 m

Hence wavelength = 10 m

4 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

Read 2 more answers

In which situation is the acceleration of the car negative?

VikaD [51]

When a car is slowing down, it has a negative acceleration. Although it is not going a negative speed, it is decreasing in velocity, which is the definition of a negative acceleration.

Hope this helps!

4 0

2 years ago

Read 2 more answers

Two circles are drawn in a 12-inch by 14-inch rectangle. Each circle has a diameter of 6 inches. If the circles do not extend be

Dennis_Churaev [7]

Answer:

d = 10 inch

Explanation:

The farthest distance between the centers, is along the diagonal of the rectangle. Therefore, we need to calculate the diagonal of the rectangle, but counting the fact that we have both circles.

So if, one side is 12 inch, and the other is 14 inch, we can use the Pitagoras theorem which is:

d = √(a²) + (b)²

Where a and b, are the lenght of the rectangle, but without the lenght of the diameter of both circles.

With this, the expression is this:

d = √(14 - 6)² + (12 - 6)²

d = √64+36

d = √100

d = 10 inches

6 0

3 years ago