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natita [175]
3 years ago
11

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the fre

quency as 3 hertz, which statement about the wave is accurate?
The wave has traveled 32.4 cm in 3 seconds.
The wave has traveled 32.4 cm in 9 seconds.
The wave has traveled 97.2 cm in 3 seconds.
The wave has traveled 97.2 cm in 1 second.
Physics
1 answer:
liq [111]3 years ago
7 0

Answer:  D. the wave has traveled 97.2 cm in 1 second.

Explanation:

Got it right on edge, hope this helps :)

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A spacecraft orbits the Earth at height of 1600km. Calculate the escape velocity for the spacecraft.
guapka [62]

Answer:

what do u need help with

Explanation:

Calculate the escape velocity for the spacecraft. [G= 6.67×10^-11Nm^2kg^-2, mass of the Earth= 5.97×10^24kg, radius of the Earth= ...

7 0
3 years ago
A car achieves a velocity 50 meters per second after 5 seconds. What is the Car's acceleration?
Reptile [31]

Answer: 10 m/s

Explanation: Velocity/Time

50/5= 10

:)

7 0
3 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
4 years ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
I need help!! i’ll give 20 points
larisa86 [58]

Answer:

780 m to travel north

Explanation:

6 m over = 750

53 degree so it will take about 2 min to reach the destination

8 0
3 years ago
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