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MA_775_DIABLO [31]
3 years ago
15

A bird flies due north with a speed of 15 m/s relative to the air. The air moves due west with a speed of 11 m/s relative to the

ground. What is the robin's speed relative to the ground?
Physics
1 answer:
Aleksandr [31]3 years ago
4 0

Answer:

I'm not sure if the directions affect the speed tho

but I think it's 4? cause it's the speed relative to the ground

hope this helps:))

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A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of n
andreyandreev [35.5K]

Answer:

5.23km/s

Explanation:

Given

Radius of Earth = 6.37 * 10^6 m

Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m

Gravity Acceleration on Satellite Altitude = 1.87965m/s²

For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

Centripetal Acceleration = V²/R

So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R

Make V the subject of formula

A = V²/R

V² = AR

V = √AR

Where R = (radius of earth) + (altitude of satellite)

R = 6.37 * 10^6 + 8.2 * 10^6

R = 14.57 * 10^6m

A = 1.87965m/s²

V = √(1.87965 * 14.57x10^6)

V = √27386500.5

V = 5233.211299001789

V = 5233.2113 m/s ------- Approximated

V = 5.23km/s

7 0
3 years ago
A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it
daser333 [38]

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

T = FR = 18 * 2.4 = 43.2 Nm

According to Newton 2nd law, the angular acceleration would be

\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2

It starts from rest, then after 15s it would achieve a speed of

\omega = \alpha t = 0.021 * 15 = 0.31 rad/s

(b) The distance angle swept by it is:

\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad

Hence the work by the child

W = T\theta = 43.2 *2.314  \approx 100 J

c) Average power to work per time unit

P = \frac{W}{t} = \frac{100}{15} = 6.67 W

7 0
3 years ago
The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2
allsm [11]

Answer:

Looks like you have:

a = -.324 cos 2.5 t

In this case   ω^2 A = .324

ω = 2.5

f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec

5 0
2 years ago
The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th
DaniilM [7]

Answer:

\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}

Explanation:

The vertical component of the initial velocities are

v_v = v_0sin\theta

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

v_vt - gt^2/2 = s = 0

t(v_v - gt/2) = 0

v_v - gt/2 = 0

t = 2v_v/g = 2v_0sin\theta/g

So the ratio of the times of the flights is

t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}

8 0
3 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

v_f=1.717 m/s        

8 0
3 years ago
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