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Lena [83]
3 years ago
8

Un bloque de 25 n se encuentra suspendido por un hilo al techo vitamina la tensión que aparece en el hilo

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
7 0

Answer:

T = 25 N

Explanation:

The question says that "A 25 n block is suspended by a wire from the ceiling vitamin the tension that appears in the wire ?"

Weight of the block, W = 25 N

Weight of a body acts in downward direction and tension acts in upward direction. It would mean that,

Tension = weight of the block

T = mg

T = 25 N

Hence, the tension in the wire is 25 N.

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Does the planet exert a torque on the meteoroid with respect to the center of mass of the planet? Why or why not?
ra1l [238]

Answer:

yes

Explanation:

because the planet exerts a centripetal force on the meteorold

4 0
3 years ago
a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to
Natalka [10]

Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

4 0
3 years ago
4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of
Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle (\Phi) = 20°

Pinion speed (n_p ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear (N_G) = 50

Teeth on pinion (N_p) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = \frac{N}{P}

=\frac{20}{5}

= 4 in

From the values of Lewis Form Factor Y for (n_p ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

V = \frac{\pi d n_p}{12}

V = \frac{\pi *4*2000}{12}

V = 2094.40 ft/min

For cut or milled profile; the velocity factor (K_v) can be determined as follows:

K_v = \frac{2000+V}{2000}

K_v = \frac{2000+2094.40}{2000}

= 2.0472

However, there is need to get the value of the tangential load(W^t), in order to achieve that, we have the following expression

W^t=\frac{T}{\frac{d}{2} }

W^t = \frac{63025*H}{\frac{n_pd}{2}}

W^t = \frac{63025*30}{2000*\frac{4}{2}}

W^t = 472.69 lbf

Finally, the bending stress is calculated via the formula:

\sigma = \frac{K_vW^tp}{FY}

\sigma = \frac{2.0472*472.69*5}{1*0.321}

\sigma = 15073.07 psi

\sigma = 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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