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andriy [413]
3 years ago
9

What is the tension in the cord after the system is released from rest? Both masses (A and B) are 10-kg.

Physics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

98 N.

Explanation:

Given data: mass= 10 kg,      gravity= 9.8 m/s2

required: tension in the cord=  ?

solution:

formula of tension= mass x gravity

by putting values of mass and gravity, we get

tension= 10 x 9.8

tension= 98 N.  <u>Ans</u>

If the mass of the object which is attached with both ends of cord is 10 kg, so the tension which is a opposite force of weight is 98 N.

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Pressure is about 1000 hPa at sea level and about 500 hPa at an altitude of 5.5 km. Why doesn’t this vertical pressure gradient
saul85 [17]

Answer:

A. The upward pressure gradient force is balanced by gravity.

Explanation:

A. is correct because the pressure difference is actually generated by gravity. As in the following formula for the pressure at different points:

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B is incorrect because friction in air is too small to make an effect.

C is incorrect because the Coriolis force is horizontal, not vertical.

D is incorrect because a difference of 500 hPa = 50000 Pa, this is half of the atmospheric pressure.

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What would happen to Earth if ocean floor were created at divergent boundaries at a faster rate than it is destroyed at converge
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3 years ago
A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.
strojnjashka [21]

Answer:

D=99.4665307m \approx 99.5m

Explanation:

From the question we are told that

Mass  m=6kg

Velocity of mass  V_m=16

Force of Tunnel  F_t=8N

Length of Tunnel L_t=1.6

Height of frictional incline H_i=2.9

Angle of inclination  \angle =16 \textdegree

Acceleration due to gravity  g=9.8m/s^2

First Frictional surface has a coefficient  \alpha_1 =0.21\ for\ d_c=1

Second Frictional surface has a coefficient \alpha _2=0.1

Generally the initial Kinetic energy is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}(6)(16)^2

K.E=768

Generally the work done by the Tunnel is mathematically given as

w_t=F_t*d_t

w_t=8*1.6

w_t=12.8J

Therefore

Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t

E_t=K.E+E_t\\E_t=768J+12.8J

E_t=780.8J

Generally the energy lost while climbing is mathematically given as

E_c=mgh

E_c=(6)(9.8)(2.9)

E_c=170.52J

Generally the energy lost to friction is mathematically given as

E_f=\alpha *m*g*cos\textdegree*d_c

E_f=0.21*6*9.8*cos16*1

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Generally the energy left in the form of mass Em is mathematically given as

E_m=E_t+E_c+E_f

E_m=(768J)-(170.52)-(12)

E_m=585.48J

Since

E_m=\alpha_2*g*m*d

Therefore

It slide along the second frictional region

D=\frac{585.46}{0.1*9.81*6}

D=99.4665307m \approx 99.5m

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Answer:

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