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Semmy [17]
3 years ago
13

A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m/s. After 1.1 minutes, the finch ti

res of the tortoise's slow pace, and takes flight in the same direction for another 1.5 minutes at 11 m/s. Part A What was the average speed of the finch for this 2.6-minute interval
Physics
1 answer:
Romashka [77]3 years ago
4 0

Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance =  990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

<u>Average Speed = 6.37 m/s</u>

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6 0
3 years ago
Một chất điểm khối lượng m=200g chuyển
MrMuchimi
Can you please translate to English?
8 0
2 years ago
The total units by an objects as it changes position is called _____ ?
Alika [10]
Change in position of object = Displacment
7 0
3 years ago
A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate
qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg =  0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0
3 years ago
A 3858 N piano is to be pushed up a(n) 3.49 m frictionless plank that makes an angle of 31.6 ◦ with the horizontal. Calculate th
uranmaximum [27]
The work done will be equal to the potential energy of the piano at the final position

P.E=m.g.h

.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
.h be the side opposite to the angle x (h is the final height of the piano)
.let L be the length of the plank

sinx=opposite side / hypotenuse
= h/L

then h=L.sinx=3.49×sin31.6°=0.638m

weight w=m.g
m=w/g=3858/10=385.8kg

Consider Gravity g=10m/s2

then P.E.=m.g.h=385.8kg×10×0.638=2461.404J

then Work W=P.E.=2451.404J
8 0
3 years ago
Read 2 more answers
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