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Likurg_2 [28]
3 years ago
7

Consider three properties of the net force on an object: a magnitude, direction, and angle with respect to the objects path. For

the following situation select the choice that indicates wether these quantities are constant or changing during the course of the motion.
1. The magnitude and the angle of the net force with respect to path are constant, but the direction is changing

2. The direction and the angle of the net force with respect to path are constant, but the magnitude is changing.

3. The magnitude and direction of the net force are changing, but the angle of the net force with respect to path is constant.

4. The magnitude, direction, and angle with respect to path for the net force are constant.

5. The magnitude and direction of the net force are constant, but the angle of the net force with respect to path is changing.
Physics
1 answer:
zalisa [80]3 years ago
4 0

Answer:

Explanation:

yeah i dont know i need the answer too

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A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of lengt
damaskus [11]

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure,  41.0  PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be  14.6959  P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂    ⇒    V₂ = P₁*V₁ / P₂

where  

1  and  2  are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

P₁ = 14.6959  P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒  V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A  = (13.2985 in³) / (3.00 in²)

⇒  h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

4 0
3 years ago
Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh
GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=N_1-mg=N_1-690

Now, from Newton's second law:

F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=mg-N_2=690-N_2

Now, from Newton's second law:

F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

Difference=N_1-N_2=862.5-517.5=345\ N

Therefore, the difference between the up and down scale readings is 345 N.

4 0
3 years ago
A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the
Shkiper50 [21]

Answer:

 ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round  = 144 Kg

angle =  33.2°

moment of inertia

I = \dfrac{1}{2} m R^2

I = \dfrac{1}{2}\times 144 \times 2.75^2

    I = 544.5 kg.m²

torque = force  x radius

τ = 37.7 x  2.75

τ = 103.675 N.m

angular acceleration

\alpha= \dfrac{\tau}{I}

\alpha= \dfrac{103.675}{544.5}

 α = 0.190 rad/s²

now ,

distance = 33.2\times \dfrca{2\pi}{360}

d = 0.579 rad

we know,

using equation of rotational motion

d = \omega t + \dfrac{1}{2}\alpha t^2

0.579 = \dfrac{1}{2}\times 0.190\times t^2

 t = 2.46 s

angular speed

 ω =  α  x t

 ω = 0.19 x 2.46

 ω = 0.467 rad/s

7 0
3 years ago
The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the
kiruha [24]

Answer:

Gravitational force

Explanation:

If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.

It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.  

Hence, the correct option is (E) "Gravitational force".                        

8 0
3 years ago
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
4 years ago
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