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3 years ago

What is a truck's acceleration if the net external force on the truck is 560 N and it has a total of mass of 1000 kg?

Physics

1 answer:

lakkis [162]3 years ago

7 0

Answer:

Acceleration of Truck = 0.56 m/s²

Explanation:

Given information:

Mass of truck = 1,000 kilogram

Net force applied by Truck = 560 N

Find:

Acceleration of Truck

Computation:

Force = Mass x Acceleration

Net force applied by Truck = Mass of truck x Acceleration of Truck

560 = 1,000 x Acceleration of Truck

Acceleration of Truck = 560 / 1,000

Acceleration of Truck = 0.56 m/s²

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A potter's wheel has the shape of a solid uniform disk of mass 13.0 kg and radius 1.25 m. It spins about an axis perpendicular t

Fittoniya [83]

Answer: $10.82 kg-m^2$

Explanation:

Given

Mass of solid uniform disk $M=13 kg$

radius of disk $r=1.25 m$

mass of lump $m=1.7 kg$

distance of lump from axis $r_0=0.63$

Moment of inertia is the distribution of mass from the axis of rotation

Initial moment of inertia of disk $I_1=\frac{Mr^2}{2}$

$I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2$

Final moment of inertia $I_f$ =Moment of inertia of disk+moment of inertia of lump about axis

$I_f=\frac{Mr^2}{2}+mr_0^2$

$I_f=10.15+1.7\times 0.63^2$

$I_f=10.15+0.674$

$I_f=10.82 kg-m^2$

3 0

4 years ago

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)

$T-P_{2}=m_{2}a$ (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the same tension of the string </h2>

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Alja [10]

Answer:

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3 years ago

Birdman is flying horizontally at a

zavuch27 [327]

Answer:

X=92.49 m

Explanation:

Given that

u= 21 m/s

h= 97 m

Time taken to cover vertical distance h

h= 1/2 g t²

By putting the values

97 = 1/2 x 10 x t² ( g = 10 m/s²)

t= 4.4 s

The horizontal distance

X= u .t

X= 21 x 4.4

X=92.49 m

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4 years ago

A cannon is shot from a boat at 185 m/s at a 52.2 angle. How far away does it land?

PilotLPTM [1.2K]

Answer:

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Explanation:

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