Ask question

Ask question

All categories

2 years ago

12

A farm tractor tows a 3300-kg trailer up a 14" incline with a steady speed of 2.8 m/s. what force does the tractor exert on the

trailer ( ignore friction.)

1 answer:

gladu [14]2 years ago

7 0

The force exerted by gravity is:

F = m g

F = 3300 kg * 9.8 m/s^2

F = 32,430 N

The force exerted due to the inclined plane is:

F tractor = 32,430 N * sin 14

<span>F tractor = 7,823.75 N (answer)</span>

You might be interested in

Which equation is equivalent to Log (26•35) ?

Nady [450]

Answer:

Log (26) + Log (35)

6 0

2 years ago

What does the power of a machine measure?

GarryVolchara [31]

The power of a machine is the work/time ratio for that particular machine

Its the rate of doing work.

5 0

2 years ago

Read 2 more answers

How do scientific tests help determine the properties of substances

tino4ka555 [31]

It can help determine substances that appear similar but react differently under the same circumstances.

7 0

2 years ago

QUICK SOMEONE PLEASE HELP!!!! I’LL MARK BRAINLIEST!!!

Brilliant_brown [7]

Temperatures above 100 are gases

8 0

2 years ago

A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

mezya [45]

Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

So

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

So

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

$m = 3454.08 g$

$m = 3.45408 kg$

8 0

2 years ago