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3 years ago

5

An influence that can deform a flexible abject or change the motion of an object with mass.

1 answer:

algol133 years ago

7 0

Answer:

An influence that can deform a flexible abject or change the motion of an object with mass. Answer is A force

xXxAnimexXx

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A refrigerator is 1.8m tall, lm wide,and 0.8m deep.The center of mass is lm from the bottom, 0.5m from the side, and 0.6m from t

VikaD [51]

Answer:

F = 520 N

Explanation:

For this exercise the rotational equilibrium equation should be used

Σ τ = 0

Let's set a reference system with the origin at the back of the refrigerator and the counterclockwise rotation as positive. On the x-axis it is horizontal directed outward, eg the horizontal y-axis directed to the side and the z-axis vertical

Torque is

τ = F x r

the bold indicate vectors, we analyze each force

the applied force is horizontal along the -x axis, the arm (perpendicular distance) is directed in the z axis,

The weight of the body is the vertical direction of the z-axis, so the arm is on the x-axis

-F z + W x = 0

F z = W x

F = $\frac{x}{z}$ W

The exercise indicates the point of application of the force z = 1.5 m and the weight is placed in the center of mass of the body x = 0.6 m, we are assuming that the force is applied in the wide center of the refrigerator

let's calculate

F = 1300 0.6 / 1.5

F = 520 N

5 0

2 years ago

a typical top fuel dragstar has a mass of 900 kg what is the force applied to a dragstar by the static friction of the strip the

AnnZ [28]

Wonderful.
You've given us ' m '.
Now, if we only had ' a ', we could come back at you with ' f '.

6 0

3 years ago

If the fly experienced a force of 100 N when it hits how big was the force on the vehicle in this collision

iogann1982 [59]

Answer:

100N

Explanation:

because newton's third law of motion states that if body A exerts a force on the body B, then Body B will exert an equal force but opposite in direction force on body A

so if the fly experienced 100N, then the car will also experience 100N

hope you get it

please mark

8 0

3 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

2 years ago

A piece of wood with density of 824 kg/m^3 is tied to the bottom of a pool and the wood does not move. The volume of the wood is

Reil [10]

Answer:

2069.76 N

Explanation:

density of wood = 824 kg/m^3

density of water = 1000 kg/m^3

Volume of wood = 1.2 m^3

True weight of wood = volume of wood x density of wood x gravity

True weight of wood = 1.2 x 824 x 9.8 = 9690.24 N

Buoyant force acting on wood = volume of wood x density of water x gravity

Buoyant force acting on wood = 1.2 x 1000 x 9.8 = 11760 N

Tension in the rope = Buoyant force - True weight

tension in rope = 11760 - 9690.24 = 2069.76 N

8 0

2 years ago