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belka [17]
3 years ago
12

In order to be _______________ forces, their effects must cancel each other out and not cause a change in the object's motion

Physics
1 answer:
Tju [1.3M]3 years ago
4 0
Balancing.

When the same force is applied from both sides,the forces cancel out each other.
You might be interested in
Please help <br> Physics is so confusing
crimeas [40]

Answer:

below

Explanation:

sin a = 4/5

a = 53.1

tan theta = 3/4

theta = 36.89

4 0
2 years ago
( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration
Talja [164]

Answer:

4.725 m/s^{2}

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then a=\frac {F}{m}  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula m=\frac {F}{a}

For m1= \frac {F}{13.5} and m2=\frac {F}{3.5} hence F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}

4 0
3 years ago
Show solution for # 4
velikii [3]
M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).

The momentum before colision is equal with the momentum after colision:

m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem

=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
7 0
3 years ago
2 pts
const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0
2 years ago
A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

V =\pi D( \frac{N}{60} )

V = 3.14 (0.034) \frac{(1650)}{60}

V = 2.93 \frac{m}{s}

Form factor for the pinion gear is

Y = 0.303

Now

K_{v} = \frac{6.1 +0.303}{6.1} = 1.049

Force on gear tooth

F = \frac{P}{V}

F = \frac{1200}{2.93}

F = 408.73 N

Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

\sigma = 35.38 M pa

This is the value of bending stress on the pinion

8 0
3 years ago
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