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Levart [38]
3 years ago
10

Which of the following is true A. If the sum of the external forces on an object is zero, then the object must be in equilibrium

. B. If the sum of the external torques on an object is zero, then the sum of the external forces on it must also be zero. C. If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero. D. If an object remains at rest, then the sum of both the external torques and the external forces on the object must be zero. E. If the sum of both the external torques and the external forces on an object is zero, then the object must be at rest.
Physics
1 answer:
yanalaym [24]3 years ago
3 0

Answer:

A. If the sum of the external forces on an object is zero, then the object must be in equilibrium

Explanation:

Equilibrium, in physics, the condition of a system when neither its state of motion nor its internal energy state tends to change with time.

For a single particle, equilibrium arises if the vector sum of all forces acting upon the particle is zero.

the object is at equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton.

There are three types of equilibrium: stable, unstable, and neutral

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The answer is 2,130,000

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3 years ago
A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are ????s = 0.9 an
Xelga [282]

Answer:

Maximum acceleration of the truck is 5.25 m/s^2

Explanation:

To find the maximum acceleration the truck first we need to calculate friction between truck and wood. Because of wood is not moving, coefficient of c_{s} need to be used for it. Then the formula will be:

F_{s}=c_{s}*200*9.8*cos(20)\\F_{s}=0.9*200*9.8*0.94\\F_{s}=1657.62

1657.62 Newton is the friction.

So force against friction need to be at most 1657.62 N. Then equation will be:

F_{s}=F_{a}+F_{m}\\1657.62=200*9.8*sin(20)+200*a*cos(20)\\1657.62=670.36+187.94*a\\987.26=187.94*a\\a=5.25

Maximum acceleration of the truck need to be 5.25 m/s^2

4 0
3 years ago
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
mina [271]

Answer:

n = 2.06 moles

Explanation:

The absolute pressure at depth of 27 inches can be calculated by:

Pressure = Pressure read + Zero Gauge pressure

Zero Gauge pressure = 14.7 psi

Pressure read = 480 psi

Total pressure = 480 psi + 14.7 psi = 494.7 psi

P (psi) = 1/14.696  P(atm)

So, Pressure = 33.66 atm

Temperature = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

T = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

33.66 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 2.06 moles

7 0
3 years ago
For a spherical mirror, the focal length is equal to ____ the radius of curvature of the mirror.
maksim [4K]
The focal length is equal to one-half the radius of curvature of the mirror.
4 0
2 years ago
If 56 grams of carbon monoxide burns in oxygen to produce 88 grams of carbon dioxide, the mass of oxygen involved in the reactio
pentagon [3]

Answer : The mass of oxygen involved in the reaction is, 32 grams.

Explanation : Given,

Mass of carbon monoxide = 56 g

Mass of carbon dioxide = 88 g

Molar mass of carbon monoxide (CO) = 28 g/mole

Molar mass of oxygen (O_2) = 32 g/mole

First we have to calculate the moles of carbon monoxide.

\text{Moles of CO}=\frac{\text{Mass of CO}}{\text{Molar mass of CO}}=\frac{56g}{28g/mole}=2moles

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction will be,

2CO+O_2\rightarrow 3CO_2

From the balanced chemical reaction, we conclude that

2 moles of CO react with 1 mole of O_2

So, the moles of O_2 = 1 mole

Now we have to calculate the mass of oxygen.

\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2

\text{Mass of }O_2=1mole\times 32g/mole=32g

Therefore, the mass of oxygen involved in the reaction is, 32 grams.

5 0
3 years ago
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