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baherus [9]
2 years ago
12

As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

of the black horse?
Physics
2 answers:
marishachu [46]2 years ago
8 0
2 min = 120 sec

120/15 = 8

The black horse represents 8 seconds.
Wewaii [24]2 years ago
5 0

Answer:

8 seconds

Explanation:

The time taken by the merry go round to complete one round is called time period.

Number of rounds = 15

Time taken = 2 minutes = 2 x 60 = 120 second

15 rounds are completetd in 120 seconds

1 round completed in 120 / 15 = 8 seconds

So, the time period is 8 seconds.

You might be interested in
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0
2 years ago
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P
Ber [7]

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

5 0
2 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
1 year ago
You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
5 0
2 years ago
You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.
ankoles [38]

Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

Mass = 60 Kg

Energy = 5000 J

We need to calculate the specific heat

Using formula of energy

Q= mc\Delta T

c =\dfrac{Q}{m\Delta T}

Put the value into the formula

c=\dfrac{5000}{60\times(37-13)}

c=3.47222\ J/kg^{\circ}C

Hence, The specific heat is 3.47222 J/kg°C.

5 0
1 year ago
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