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hoa [83]
3 years ago
5

3. What does the difference in force depend on?

Physics
2 answers:
dangina [55]3 years ago
4 0

Answer:

it depends on the relative masses of the objects.

Explanation:

zvonat [6]3 years ago
3 0
It depends on the mass of an object
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A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t
weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

\theta=90^{\circ}

We have to find the their velocity immediately after the tackle.

Initial momentum,P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s

According to law of conservation of momentum

Initial momentum=Final momentum=(m+m')V

627.6=(85+90)V=175V

V=\frac{627.6}{175}=3.59 m/s

3 0
3 years ago
According to Newton's 3rd Law, an object's momentum depends on it's velocity and mass. 4 dog-sled teams competed to see who coul
podryga [215]

Answer:

C.) Sled Team C 28 kg moving at 12m/s

I'm pretty sure.

7 0
3 years ago
You are investigating an elevator accident which happened in a tall building. An elevator in this building is attached to a stro
Advocard [28]

Answer:

a) F = 2250 Ib

b) F = 550 Ib

c) new max force ( F newmax ) = 2850 Ib

Explanation:

A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up

max capacity  of elevator = 24000 Ibs

counterweight = 1000 Ibs

To calculate the force (F) :

we first calculate the Tension using this relationship

Counterweight (1000) - T =  ( 1000 / g ) ( g/4 )

Hence T = 750 Ib

next determine F

750 + F - 2400 = 2400 / 4

hence F = 2250 Ib

B ) calculate Tension first

T - 1000 = ( 1000/g ) ( g/4)

T = 1250 Ib

F = 2400 -1250 - 2400/ 4

F = 550 Ib

C ) determine design limit

Max = 2400 * 1.2 = 2880 Ib

750 + new force - 2880 = 2880 / 4

new max force ( F newmax ) = 2850 Ib

8 0
3 years ago
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang
koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × 10^{-4}  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = \int \vec{B}.d\vec{A}   .................1

and Ф = BA cos∅    ............2

so B = \frac{\phi }{Acos\theta }    

and we know

A = ab

so

B = \frac{\phi }{abcos\theta }    ..............3

put here value

B =  \frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}  

solve we get

B = 0.692 T

8 0
3 years ago
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
Mumz [18]

Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

3 0
3 years ago
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