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andreev551 [17]

3 years ago

14

A 5.0-m/s wind from the west gives the bird a groundspeed of 15 m/s flying east and 5 m/s flying west. Find the time for the rou

nd-trip flight under these conditions.

1 answer:

Andreyy893 years ago

6 0

The bird's flying speed through the air is 10 m/s. To answer the question, we need to know the facts of the flight.

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17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set

lapo4ka [179]

Answer:

0.30581

0.24464

Explanation:

$\mu_s$ = Coefficient of static friction

$\mu_k$ = Coefficient of kinetic friction

$F_f$ = 75 N

$F_k$ = 60 N

Normal force

$F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N$

Frictional force

$F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581$

The coefficient of static friction is 0.30581

Kinetic force

$F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464$

The coefficient of kinetic friction is 0.24464

4 0

2 years ago

A star with a mass like the Sun which will soon die is observed to be surrounded by a large amount of dust and gas -- all materi

Dafna11 [192]

Answer: D. Infrared

Infrared is the best way to observe it.

5 0

2 years ago

The best type of exercise for weight loss is

lbvjy [14]

Answer:

finding a workout.

Explanation:

6 0

2 years ago

a man pushes on a wall with a force of 50N. what are the size and direction of the force that the wall exert on the man?

Ira Lisetskai [31]

According to Newton's 3rd law of motion, every action has an equal and opposite reaction. So Wall must apply a force of 50 Newton but in opposite direction. Hope you got it :-D

5 0

2 years ago

Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.

-The equations of motion from Newton's Second Law are:

$mBg-T=mBa$ where a is the acceleration.

#Substituting for $T$ (tension) gives:

$mBg-mAsin\theta-mAa=mBa$

#and solving for $a:$

$a=\frac{mBg-mAgSin\theta}{mA+mB}$ which is the system's acceleration.

8 0

3 years ago