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3 years ago

8

The moon’s rotational period and revolution period are equal. What is the result of this? Enter your answer in the space provide

d.

1 answer:

beks73 [17]3 years ago

5 0

Answer The Moon has synchronous rotation: it's rotation period is the same as its period of revolution

Explanation:

The Moon has synchronous rotation: it's rotation period is the same as its period of revolution

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An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk

ahrayia [7]

Answer:

C) 1.0 m

Explanation:

The component of the velocity parallel to the sidewalk is:

vₓ = v cos θ

vₓ = 0.1 m/s cos 45°

vₓ = 0.0707 m/s

The distance traveled after 14 seconds is:

d = vₓ t

d = (0.0707 m/s) (14 s)

d = 0.99 m

Closest answer is C) 1.0 m.

6 0

2 years ago

Read 2 more answers

What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio

pantera1 [17]

If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),

where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.

3 0

2 years ago

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

3 years ago

Suppose that a pendulum has a period of

Yakvenalex [24]

Explanation:

period of pendulum = time taken for 1 oscillation = time taken for 1 complete back and forth vibration

q1 ans is given in question its 1.5 sec

q2 ans is 1.5 sec longer than 1 sec period

4 0

2 years ago

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

iren [92.7K]

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

$U=\frac{1}{2}kx^2$

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

$K=\frac{1}{2}mv^2$

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

$\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s$

4 0

3 years ago