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1 year ago

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The free-body diagram below represents the force of several vehicles driving across a bridge. Assume that the bridge is in stati

c equilibrium and that it has zero weight, and solve for the unknown reaction force.

1 answer:

krek1111 [17]1 year ago

7 0

In order to calculate the unknown reaction force, we need to know that the sum of forces pointing down is equal the sum of the forces pointing up, so all forces will be in equilibrium.

So we have:

$\begin{gathered} 17800+150000+13200=121000+x \\ 181000=121000+x \\ x=181000-121000 \\ x=60000 \end{gathered}$

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A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm

Rus_ich [418]

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

v = d / t

t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

t = 150/343

t = 0.437 s

This same sound takes much longer to reach you

t₂ = 127 10³/343

t₂ = 370 s

6 0

3 years ago

Which of the following concepts best explains why a mass of low-density material in the mantle rises?Multiple Choice Strain Buoy

SOVA2 [1]

Answer:

Bouyancy

Explanation:

Bouyancy occurs when the upthrust exerted on an object is equal to the weight of object displaced. It is mostly applicable to low density objects for example balloon. When balloon is displaced in water, it floats. This is due to the effect of the upthrust acting on the balloon which allows the balloon to float and which is opposite the weight.

Note that the weight acts downwards the object while the upthrust always acts opposite (upward)

8 0

3 years ago

Which is more dense, cold freshwater or warm seawater?

svp [43]

Cold freshwater<span> is </span>denser<span> than </span>warm seawater<span>, because of the salinity and temperature variations. Cold water would have less salt since the solubility of the salt is lower as compared to warm water. Hope this answers the question. Have a nice day.</span>

4 0

3 years ago

Read 2 more answers

Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di

Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, $u=20m/s$

And the mass of the car is, $m=1000 kg$

The total distance covered by the car before stop, $s=45m$

And the final speed of the car is, $u=0m/s$

Now initial kinetic energy is,

$KE_{i}=\frac{1}{2}mu^{2}$

Substitute the value of u and m in the above equation, we get

$KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J$

Now final kinetic energy is,

$KE_{f}=\frac{1}{2}mv^{2}$

Substitute the value of v and m in the above equation, we get

$KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J$

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

$F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN$

Here, the force is negative because the force and acceleration in the opposite direction.

6 0

3 years ago