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4 years ago

Since the Wagon is being pulled down hill is it increasing (C)

Physics

1 answer:

Ket [755]4 years ago

4 0

Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.
You are correct! **

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I need help ASAP I need to get this right plz plz plz!!!!!

zhenek [66]

Answer:

option d and b..............

3 0

3 years ago

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

erma4kov [3.2K]

Answer:

c. 981 watts

$P=981\ W$

Explanation:

Given:

horizontal speed of treadmill, $v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}$

weight carried, $w=588.6\ N$

grade of the treadmill, $G\%=10\%$

Now the power can be given by:

$P=v.w$

$P=588.6\times\frac{5}{3}$ (where grade is the rise of the front edge per 100 m of the horizontal length)

$P=981\ W$

6 0

4 years ago

What is the force if the mass is 75kg and the acceleration is 24.5m/s^2

Snowcat [4.5K]

Given;

mass m = 75 kg

acceleration a = 24.5 ms²

F = ma 

F = 75 kg * 24.5 ms²

= 1837.5 kg ms².

4 0

3 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

4 years ago

The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pu

Kisachek [45]

Answer:

$a=2.36\ m/s^2$

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

$m_1g-m_2g=(m_1+m_2)a$

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

$a=2.36\ m/s^2$

Lets tension in string is T

$m_1g-T=m_1a$

$T=m_1(g-a)$

T=21.1(9.81-2.36) N

T=157.06 N

6 0

3 years ago