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drek231 [11]
3 years ago
13

A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the car from the deck of a cargo sh

ip. What is the acceleration of the car during this time?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer: 1.2m/s^2

Explanation: the force exerted on the car is 900N upwards

The mass of the car is 750kg

According to Newton's third law acceleration is proportional to force

F = ma

900 = 750a

a = 900/750

a = 1.2m/s^2

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Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used
natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is  T =8391.6 N

Explanation:

From the question we are told that

       The strain on the strain on the head is \Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m

         The contact area is  A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2  

Looking at the first diagram

           At  600 MPa of stress

               The strain is  0.3mm/mm

          At   450 MPa of stress

                 The strain is   0.0015 mm/mm

 To find the stress at  \Delta l we use the interpolation method

            \frac{\sigma_{\Delta l} -  \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l }  - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}

Substituting values

              \frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}

            \sigma _{\Delta l} -450 = 49.50

             \sigma _{\Delta l} = 499.50 MPa

Generally the force on each head is mathematically represented as

              F = \sigma_{\Delta l} * A

Substituting values

             F = 499.50*10^{6} * 2.8*10^{-6}

                =1398.6N

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

              T = 6 * F

                  = 6* 1398.6

              T =8391.6 N

                 

     

6 0
3 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
Recycled plastics cannot be used to produce new food containers because ___
Lena [83]

they cannot be sterilized

4 0
3 years ago
Electrons and protons are basic particles (and together make atoms) that have charge. What is the electric charge of an electron
trapecia [35]

Answer:

hi, here's the answer

Explanation:

the electric charge of an electron is negative (-ve).

pls mark this as the brainliest...

8 0
3 years ago
An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Alex Ar [27]

Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

            y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

            y = 77.74 10⁻⁵ m

3 0
3 years ago
Read 2 more answers
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