Ewww you might use this element to kiss someone

A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

svlad2 [7]

Force=mass*acceleration
F=ma
F=25*5
F=100 N

8 0

3 years ago

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

5 0

4 years ago

What is meant by pressure?write its unit.

Mars2501 [29]

Answer:

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed

Unit: Pascal (Pa)

1Pa = 1N/m^2

5 0

3 years ago

What happens to a muscle when you exercise it?

NISA [10]

It grows... as expected of a muscle...

5 0

3 years ago

Read 2 more answers

Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m

Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

$A=\frac{1-V_{1}^{2} }{E_{1} }-\frac{1-V_{2}^{2} }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}$

a) The maximum contact pressure is:

$P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} }) }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa$

The width of contact is:

$b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} }) } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm$

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0

3 years ago