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3 years ago

A crane lifts up two boxes. Which free body diagram shows the forces acting on block A?

Physics

2 answers:

Temka [501]3 years ago

7 0

Answer:

The 3rd graph

Explanation:

A free body diagram is a diagram which shows all the forces acting on an object.

The problem asks us to find the free body diagram of block A, so we must find all the forces acting on block A.

We have 3 forces acting on block A in total:

- The force of gravity (its weight), which pushes the block downward (in the diagram, it is the force represented with $F_{gA}$

- The tension in the rope 1, which pulls block A upwards: this force is represented with $F_{T1}$

- The tension in the rope 2, due to the weight of block 2, which pulls block A downwards: this force is represented with $F_{T2}$

Based on the direction of these 3 forces, the correct diagram is the 3rd one.

tatuchka [14]3 years ago

4 0

Answer:

t he third graph

Explanation:

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Study the roller coaster diagram. Match the labels to the correct location on the roller coaster.

IRISSAK [1]

Answer:

A- Greatest Kinetic Energy

B- Increasing Potential Energy

C- Increasing Kinetic Energy

D- Greatest Potential Energy

Explanation:

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3 0

2 years ago

Read 2 more answers

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)

Alexxx [7]

Answer:

3.15m³

Explanation:

To solve this problem, let us first find the mass of the petrol from the given dimension.

Mass = density x volume

Volume of petrol = 4.2m³

Density of petrol = 0.3kgm⁻³

Mass of petrol = 4.2 x 0.3 = 1.26kg

So;

We can now find the volume of the alcohol

Volume of alcohol = $\frac{mass}{density}$

Mass of alcohol = 1.26kg

Density of alcohol = 0.4kgm⁻³

Volume of alcohol = $\frac{1.26}{0.4}$ = 3.15m³

7 0

3 years ago

What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs

Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

the quantity expressed by the body resisting angular acceleration.
It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

#SPJ4

7 0

2 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago