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Shkiper50 [21]
2 years ago
12

Can someone help with my physics homework? please

Physics
1 answer:
Murrr4er [49]2 years ago
3 0

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

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Answer:

3.1 m/s²

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Given:

Mass of the balloon (m) = 11.4 g = 0.0114 kg ( 1 kg = 1000 g)

Force acting on the balloon (F) = 0.035 N

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Now, we know that, from Newton's second law, net force acting on an object is equal to the product of its mass and acceleration.

Therefore, framing in equation form, we have:

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Rewriting in terms of acceleration 'a', we get:

a=\frac{F}{m}

Now, substitute the given values and solve for 'a'. This gives,

a=\frac{0.035\ N}{0.0114\ kg}\\\\a=3.07\approx 3.1\ m/s^2(Nearest\ tenth)

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2 years ago
A bullet is fired horizontally, and at the same instant a second bullet is dropped from the same height. Ignore air resistance.
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<u>Answer: They hit at the same time</u>

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Ann and Bob are carrying a 18.5 kg table that is 2.25 m long. A 8.33 kg box sits on the table 0.750 m from Ann. How much lift fo
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Answer:

F = 118 N

Explanation:

Assume Ann and Bob lift at their respective ends of the table

Sum moments about Bob's position to zero.

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2 years ago
If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of e
ryzh [129]

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

K = \frac{A*E}{L}K=

L

A∗E

L = \frac{A*E}{K}L=

K

A∗E

A=\frac{\pi d^{2}}{4}A=

4

πd

2

d = 20-mm = 0.02 m

A=\frac{\pi (0.02)^{2}}{4}A=

4

π(0.02)

2

A = 0.0003 m²

L = \frac{A*E}{K}L=

K

A∗E

L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=

55X10

6

(0.0003142)∗(200X10

9

)

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}σ=

A∗E

F∗L

\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=

(0.0003142)∗(200X10

9

)

(60X10

3

)∗(1.14)

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

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