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il63 [147K]
3 years ago
10

Jerome places a bag of flour on a scale. The scale shows that the bag has a weight of 17 N. Which is the reaction force of the b

ag sitting on the scale?
The scale exerts a 17 N force up on the bag.
The scale exerts a 17 N force down on the counter.
Earth exerts a 17 N force down on the bag.
The bag exerts a 17 N force down on the scale.
Physics
1 answer:
Leno4ka [110]3 years ago
3 0

Answer:

A

Explanation:

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A capacitor is charged to a potential difference of 12v it delivers 40% of its stored energy to a lamp what is the final potenti
Korolek [52]
Call the capacitance C. 
<span>Note the energy in a capacitor with voltage V is E =½CV². </span>

<span>Initial energy = ½C(12)² = 72C </span>

<span>40% of energy is delivered, so 60% remains.in the capacitor. </span>
<span>Remaining energy = (60/100) x 72C =43.2C </span>

<span>If the final potential difference is X, the energy stored is ½CX² </span>
<span>½CX² = 43.2C </span>
<span>X² = 2 x 43.2 = 86.4 </span>
<span>X = 9.3V</span>
6 0
3 years ago
Marcie shovels snow after a storm by exerting a force of 32 N on her shovel at an angle of 50 degrees to the vertical. What is t
gulaghasi [49]

Answer:

The horizontal component of the force exerted by Marcie is 25 N

Explanation:

Lets explain how to solve the question

Marcie shovels snow after a storm by exerting a force of 32 N on her

shovel at an angle of 50 degrees to the vertical

We need to find the horizontal component of the force exerted by Marci

The force direction is 50° to the vertical means

→ tan(50) = \frac{Fx}{Fy} , where x is the horizontal component

of the force and y is vertical component

→ tan(50) = \frac{sin(50)}{cos(50)}

By comparing between the two expressions of tan(50)

→ The horizontal component is Fx = F sin(50)

→ The vertical component is Fy = F cos(50)

We need the horizontal component

→ F = 32 N

→ The horizontal component = 32 sin(50) = 24.51 N ≅ 25 N

<em></em>

<em>The horizontal component of the force exerted by Marcie is 25 N</em>

6 0
3 years ago
Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?
cluponka [151]

Answer:

Torque, \tau=0i+0j-70k

Explanation:

It is given that,

Force acting on the particle, F=-10j\ N

Position of the particle, r=(7i+5j)\ m

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

\tau=r\times F

\tau=(7i+5j)\times (-10j)

The cross product of vectors is given by :

\tau=\begin{pmatrix}0&0&-70\end{pmatrix}

or

\tau=0i+0j-70k

So, the torque on the particle about the origin 0i+0j-70k. Hence, this is the required solution.

6 0
3 years ago
You and your fellow students have been challenged to design a rubber band powered car. You will use the elastic potential energy
Tcecarenko [31]

Answer:

The correct answer is A

Explanation:

3 0
3 years ago
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density
blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

6 0
3 years ago
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