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Orlov [11]
3 years ago
10

A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th

e rear wheels drive over the scale, it reads 6500 N. The distance between the front and rear wheels is 3.20 m Determine the distance between the front wheels and the truck's center of gravity.
Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

x_2=1.60m

Explanation:

From the Question We are told that

Initial Force F_1=5800N

Final Force F_2=6500N

Distance between the front and rear wheels \triangle x=3.20 m

Since

 \triangle x=3.20 m

Therefore

 x_1+x_2=3.20

 x_1=3.20-x_2

Generally the equation for The center of mass is at x_2 is mathematically

given by

 x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}

 x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}

 2*F_1*x_2 =3.20F_1

 x_2=1.60m

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A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a
saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

  • Boyle's Law,   (P₁V₁ =P₂V₂)
  • Charles' Law  (V₁/T₁ =V₂/T₂)
  • Gay-Lussac's Law.  (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) \frac{5}{278} =\frac{P_2}{313}  \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm

7 0
4 years ago
A block with a mass of 31.8 kg is pushed on a frictionless
OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

W = F.d.\cos\theta

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

W = F.d

Given:

F = 75 N

m = 31.8 kg

Final velocity v_f = 15.3 \frac{m}{s}

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\

Now we can determine the displacement from the following formula:

d = \frac{1}{2}a^2+v_0t+d_0

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

d = \frac{1}{2}at^2\\

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m

and the corresponding work:

W = F\cdot d=75 N\cdot 48.94 m =3670.5J


7 0
3 years ago
An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break
mamaluj [8]

Answer:

Value that the spring constant k = 12Mg / h

Explanation:

According to 2nd law of Newton:

upward force of the spring= F

The weight of the elevator W = mg

            F = Mg = M(5g)

==> F =6Mg.

As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence

  F =ks = 6Mg

==>   s = 6Mg/k

We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:

            Mg(h+s) = 1/2ks2

And plugging in our expression for s:

         Mg(h+6Mg/k)= 1/2k(6Mg / k)2

          gh + 6M2g2/k = 1/2k(36M2g2 /k2)

        Mgh +6M2g2/k = 1/2k(36M2g2 /k2)

             gh + 6Mg2/k = 18Mg2 / k

                             gh = 12Mg2 / k

                               h = 12Mg / k

                               k = 12Mg / h

7 0
3 years ago
Read 2 more answers
A kid is on a stationary sled, on
gregori [183]

The mass of the kid and the sled is 50.0 kg

Explanation:

The sled starts to move when the force applied becomes larger than the maximum force of static friction, which is given by

F_{max} = \mu_s mg

where

\mu_s is the coefficient of static friction

m is the mass of the kid + the sled

g=9.8 m/s^2 is the acceleration of gravity

Here the sled starts moving when the force applied is 61.2 N, so

F_{max}=61.2 N

And we also know

\mu_s = 0.125

Therefore, we can find the mass of the kid and the sled by re-arranging the equation:

m=\frac{F_{max}}{\mu_s g}=\frac{61.2}{(0.125)(9.8)}=50.0 kg

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

6 0
3 years ago
I WILL MARK AS BRAINLIST
puteri [66]

Explanation:

Area=1.5(1.5)=2.25m^2

Force of gravity=10N

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

⟼Pressure=

Area

Force

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}

⟼Pressure=

2.25

10

\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}

⟼Pressure=4.4Pa

5 0
3 years ago
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