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alukav5142 [94]
3 years ago
11

Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. Wh

at is the velocity in problems per hour?
a. 6 per hour
b. 4 per hour
c. 10 per hour
d. 0.67 per hour
e. 15 per hour
Physics
2 answers:
ioda3 years ago
4 0

Answer:

option (b) is correct

Explanation:

time t = 1 and half hour = 1.5 hour

Number of problems, n = 6

So, the velocity in problems

v = n / t

v = 6 / 1.5

v = 4

So, the rate is 4 problems per hour.

hodyreva [135]3 years ago
3 0

Answer:

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Explanation:

given data

worked on  homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour

solution

we know that Shirley solve complete 6 accounting homework problem in 1.5 hour so her velocity  in problems per hour will be as

velocity in problems per hour =  \frac{complete\ problem}{time\ taken}   ..................1

put here value we will get

velocity in problems per hour =  \frac{6}{1.5}

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

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I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

7 0
3 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
3 years ago
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damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

Now we use the equation W=force x displacement =75 x 0.3=22.5J

I hope this satisfies you. If u have any further questions please let me know.

I hope u will follow me and make this the brainliest answer.

3 0
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Leni [432]

Answer:

1: True

2: False

3: True

4: False

5: True

3 0
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