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3 years ago

E=6.63E-34J-s*4.6E14 hz

Physics

1 answer:

Studentka2010 [4]3 years ago

4 0

Que tipo de pregunta es esa menos ósea estas menos

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A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case

vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction, $\rm \mu_k$ =0.520

Coefficient of static friction, $\rm \mu_s=0.520$

The friction force, F=?

Resolve the force in the inclined plane;

$\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times cos 25 ^0 \\\\ F= 46.2 \ N$

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

#SPJ1

5 0

2 years ago

To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the

zhannawk [14.2K]

Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

$R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}$

$R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}$

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

8 0

4 years ago

Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =

Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

$v = 3.5596 \frac{m}{s}$

Explanation:

The motion have the velocity that will make the block move using:

$F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}$

$m_{1} = 1.2 Kg$

$m_{2} = 2.8 Kg$

$r = 0,45 m$

μ $s= 0,54$

Resolving:

$\frac{m_{1}*v^{2} }{r} = us*m_{1}* g + m_{2} * g$

$v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }$

$v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}$

$v^{2} = 12.6714 \frac{m^{2} }{s^{2} }$

$v = 3.559 \frac{m}{s}$

4 0

3 years ago

Help plss i just need it plss

Travka [436]

Answer:

ooh thanks but don't give me advise

4 0

3 years ago

5. A box weighs 196 N. A rope is tied to the box. What is the

natta225 [31]

Answer:

296 N

Explanation:

Draw a free body diagram. The box has two forces on it: tension up and weight down.

Apply Newton's second law:

∑F = ma

T − mg = ma

T = m (g + a)

Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:

T = (20 kg) (9.8 m/s² + 5 m/s²)

T = 296 N

5 0

3 years ago